Permutation Reso Sheet

PSCMBGB200310

PART - I : SUBJECTIVE QUESTIONS

Hkkx- I : fo"k;kRed iz'u ¼SUBJECTIVEQUESTIONS ½ ½ Section (A) : Problem based on Problems, Arrangements Arrangements of given objects/Selection of given object (PCAD/PCSD)

[k.M(A):nhxbZoLrqvksadspquko]foU;klijvk/kkfjrleL;k,¡ A-1.

How many 3 digit even numbers can be formed using the digits 1, 2, 3, 4, 5 (repetition allowed)? allowed)?

vadks1,2,3,4,5 dhlgk;rkls 3vadks dh fdruh le la[;k,¡ cukbZ bZ tk ldrh gSa ;fn vadks dh iqujko`fÙk gks ldrh gSA Sol.

Ans. 50 Total even number s (numbers whose unit digit is 2)

2 5

×

5

= 25 (numbers whose unit digit is 4)

4 5 × on adding = 50 Hindi.

5

= 25

dqy le la[;k,¡ (la[;k,¡ ftuds bdkbZ dk vad 2 gS)

2 5

×

5

= 25 (la[;k,¡ ftuds bdkbZ dk vad 4 gSA)

4 5 × tksM+usij= 50 A-2.

5

= 25

There are 10 buses operating between places A and B. In how many ways a person can go from place A to place plac e B and an d return ret urn to place pla ce A, if he retur re turns ns in a diff erent ere nt bus? bu s? nksLFkkuksa AvkSj B dschp 10 clslapkfyrgksrhgSaA,dO;fDrLFkku A ls LFkku BijtkdjokilLFkku Aij

fdrusrjhdksalsvkldrkgS;fnog okilhesanwljhcldkmi;ksxdjsaA Sol.

Ans. 90 10 × 9 = 90

A-3.

The digits from 0 to 9 are written on slips of paper and placed in a box. Four of the slips are drawn at random and placed in the order. How many out comes are possible? dkxtdhifpZ;ksaij 0 ls 9 rd vad fy[kdj ,d cDls esa Mky nh tkrh gSA cDls esa ls ;kn`fPNd :i ls 4 ifpZ;k¡

fudkydjØelsj[khtkrhgSaAlEHkkforurhtksadhla[;kfdruhgksxh\ 10

Sol.

Ans. 10P 4

P4

A-4.

Find the number of 6 digit numbers that ends with 21 (eg. 537621), without repetition of digits.

vadks dh fcuk iqujko`fÙk ds6 vadksdhfdruhla[;k,¡cukbZtkldrhgS]tks 21 lslekIrgksrhgS\ (eg.537621) Sol.

Ans. 7. 7P3 Total number of 6 digit number that ends with 2, 1 6 vadksadhdqyla[;k;satks21 lslekIrgksrhgSA

PERMUTATION & COMBINATION - 1

i.e. Hence total number of ways is 7 × 7 × 6 × 5 = 7 × 7P 3 dqy rjhdksa dh la[;k = 7 × 7 × 6 × 5 = 7 × 7P3 A-5.

A coi n is toss t ossed ed n ti mes , then the n fin d the number num ber of possi p ossible ble outc o utc omes. ome s.

,d flDds dks n ckj mNkyusij izkIr urhtksa dh la[;k fdruhgksxh\ Sol.

A-6.

2n a coin is tossed one times the number of out comes in 2. a coin is tossed two times the number of out comes in 2 2. a coin is tossed n-times the number of out comes in 2 n. tc ,d flDds dks ,d ckj mNkyk tk;s rks dqy 2 fLFkfr;k¡lEHkogS tc,dflDdsdksnksckjmNkyktk;srks dqy22 fLFkfr;k¡lEHkogS tc ,d flDds dks n-ckj mNkyk tk;s rks dqy 2n fLFkfr;k¡ lEHko gS

Ans. When When When

( i)

If

1 1 x + = , find 'x'. 9! 10 ! 11! 1

1

x

;fn 9 ! + 10 ! = 11! gks] rks 'x' dk eku Kkr dhft,A

Sol.

A-7.

( i i)

If nP5 = 42. nP3, find 'n'. ;fn nP5 = 42. nP3, gks] rks 'n' dk eku Kkr dhft,A

Ans.

( i)

(i)

10 10 !

121.



1 10 !



(ii)

x 11!



11 10 !

(ii)

n! n! = 42 . (n – 5)! (n – 3)!



(n – 3) (n – 4) = 7× 6



10

x 11.10!



x = 121



7× 6



n = 10

How many words can be forme d by using all the letters of the word 'MONDAY' if each word start with a consonant. 'MONDAY' 'kCn ds lHkh v{kjksa dk mi;ksx djrs gq, fdrus 'kCn cuk, tk ldrs gSa tcfd izR;sd 'kCn ,d O;atu (consonant)lsizkjEHkgksrkgSA Ans.

480

Sol. Hence total number of ways. = 4×25 = 4 × 120 = 480

vr% dqy rjhdksa dh la[;k = 4×25 = 4 × 120 = 480 A-8.

Sol.

Find the number of natural numbers from 1 to 1000 having none of their digits repeated. 1 ls1000 rd,slhizkd`rla[;kvksadhla[;kKkrdhft,ftuesa ,dHkhvaddhiqujko`fÙkughagksrhgS gSA 738 Ans. Number of one digit numbers = 9 Number of 2 digits numbers = 9 × 9 = 81 Number of 3 digits numbers = 9 × 9 × 8 = 648 PERMUTATION & COMBINATION - 2

total numbers = 9 + 81 + 648 = 738 Hindi. ,d vad dh la[;kvksa dh la[;k = 9 2 vadksa dh la[;kvksa dh la[;k= 9 × 9 = 81 3 vadksa dh la[;kvksa dh la[;k = 9 × 9 × 8 = 648 dqy la[;k,¡ = 9 + 81 + 648 = 738 A-9.

( i)

If nC3 = nC 5, find the value of nC2.

;fn nC3 = nC5 gks]rks nC2 dk eku Kkr dhft,A

Sol.

C3 : nC 3 = 11 11 : 1, find 'n '. 2n n ;fn C3 : C3 = 11 : 1 gks] rks 'n' dk eku Kkr dhft,A

( i i)

If

( i i i)

If

Ans. ( i)

   

2n

n–1

Cr : nC r : n + 1 C r = 6 : 9 : 13, f ind n & r. ;fn n – 1Cr : nCr : n + 1Cr = 6 : 9 : 13 gks] rks n vkSj r

( i) 28 (ii) Hence pwafdnC3 = nC5

n! 3! n – 3

=

(iii)

n = 1 2, r = 4.

n! 5! (n – 5 )!

5 × 4 × 3! × n – 5 ! = 3! × (n – 3) (n – 4) ( n – 5)! (n – 3) (n – 4) = 5 × 4  n–3=5 n= 8 Hence vr

(ii)

6

2n C

3 :

nC

%

3 =

nC

2 =

8C

2 =

87 = 28 1 2

11 : 1



2n! 3! n!– 3 11  = 3! (2n – 3)! n! 1



2n! (n – 3)! 11  = n! (2n – 3 )! 1



2. (2n – 1) . 2 (n – 1) = 11 × (n – 1) (n – 2)  8n – 4 = 11 n – 22 3n = 18

(iii)

n–1 C

n –1





n

r :

Cr

Cr

nC



r :

n+1 C

6 9

n – r 6  n 9

r =

Cr 9 again iqu% n 1 = Cr 13

n! (n – r )! (n  1) n! (n  1 – r ) (n – r )!

(n – 1)! (n – r – 1)!. (8 )! (n)! (n – r )!.(r )!





2n = 3n – 3r



n = 3r n! r ! (n – r )! (n  1)! r ! (n  1 – r )!







n= 6

6 : 9 : 13

n



ds eku Kkr dhft,A

9 13







6 9



(n – 1)! (n – r – 1)!. (8)! n(n – 1)! (n – r )!.(n – r – 1)!



6 9

9 13

2r  1 9  3r  1 13




3r  1 – r 9  3r  1 13 26 r + 13 = 27 r + 9 r=4 n = 12



2r  1 9  3r  1 13 r=4

    Section (B) : Problem based on Selection as well as arrangement of objects/Rank of word (PCSR/PCSI)

[k.M (B):nhxbZoLrqvksadspquko],oamudsfoU;klijvk/kkfjrleL;k,¡]fn;sx;s'kCndhojh;rkij vk/kkfjrleL;k,¡ B-1.

Find the number of words those can be form ed by using all letters of the word 'DAUGHTER', if all the vowels must not be together. 'DAUGHTER''kCn ds lHkh v{kjksa dk mi;ksx djrs gq;s fdrus 'kCn cuk, tk ldrs gSa tcfd lHkh Loj (vowles)

,dlkFkuvk;sa\ Sol.

Hindi.

Ans. 36000 Total number of ways is if there is number condition is 8 ! = 40320 Again if all vowels ar e toge ther i.e. A E U DGHT R. so total number of ways = 6 ! × 3 ! = 4320 Hence total number of ways if all vowels all do not together is 40320 – 4320 = 36000 ways

dqyrjhdksadhla[;ktcfddksbZizfrcU/kughafn;kx;kgks 8 ! = 40320

;fnlHkh Loj ,d lkFk gks A E U DGHT R.

dqy rjhdksa dh la[;k = 6 ! × 3 ! = 4320 vr% dqy rjhdksa dh la[;k tcfd lHkh Loj ,d lkFk u vk;saA 40320 – 4320 = 36000 ways B-2.

A number loc k has 4 dials, each dial has the digit s 0, 1, 2, ...... .., 9. What is the maxim um unsuc cessful attempts to open the lock?

,d la[;kRed rkys ds 4 Mk;y gSaA izR;sd Mk;y esa 0, 1, 2, ........, 9 rd vad gSaA rkys dks [kksyus ds vf/kdre vlQyiz;klksadhla[;kfdruhgksxh\ Sol.

Hindi.

Ans. 9999 Total unsuccessful attempts to open the lock is = (total attempt – successful attempt) = 10 × 10 × 10 × 10 – 1 = 10000 – 1 = 9999

rkys dks [kksyus dh dqy vlQy iz;kl = (dqy iz;kl & lQy iz;kl) = 10 × 10 × 10 × 10 – 1 = 10000 – 1 = 9999

B-3.

If all the letters of the word 'AGAIN' are a rranged in all possible ways & put in dictionary order, what is the 50 th word? ;fn'AGAIN''kCndslHkhv{kjksadkslHkhlEHkorjhdksalsO;ofLFkrdj'kCndks"kdsØeesaj[kktk;s]rks50ok¡'kCn

dkSulkgksxk\ Ans. Sol.

NAAIG = 24 ways rjhds

= 12 ways rjhds


= 12 ways rjhds

and here ;gka 50th

B-4.

NAAIG

How many different permutations are possible using all the letters of the word MISSISSIPPI, if no two 's are together? 'kCn MISSISSIPPI ds lHkh v{kjksa dkmi;ksx djrs gq;s dqy fdrus 'kCn cuk, tk ldrs gSa] ;fn dksbZ Hkh nks 

lkFk&lkFkughavk;sa\ Sol.

Ans. 7350 Total no. of M are = 1 Total no. o f  are = 4 Total no. of P are = 2 Total no. of S are = 4 First we arrange all the words other than 's are

7! 2! 4!



765 1 2

 105

Now, there are 8 places which can be fulfilled by I's i.e. the number of ways is 8C4 105  8  7  6  5 Total required no. = 105 × 8C4 = 1 2  3  4 = 105 × 70 = 7350 Hindi. M dh dqy la[;k = 1  dh dqy la[;k= 4 P dh dqy la[;k= 2 S dh dqy la[;k= 4 igysgeI dksNksM+djlHkhv{kjksadksO;ofLFkrdjrsgSaA

7! 2! 4!



765 1 2

 105

vc8 LFkkucpsaxsatksfd }kjkHkjstkldrsgSabldsrjhdksadhla[;k 8C4 dqy vHkh"V 'kCnksa dh la[;k = 105 × 8C4 =

105  8  7  6  5 1 2  3  4

= 105 × 70 = 7350 B-5.

Six persons meet in a r oom and each shakes hands with all the others. Find the total number of hand shakes that took place. 6 O;fDr,ddejsesafeyrsgSvkSj,dnwljslsgkFkfeykrsgSa]rksgkFkfeyusdhizfØ;kdhdqyla[;kfdruhgksxh\ Ans. 15.

Sol.

gk¡Fk feykus dh izfØ;k dh dqy la[;k = 6C2 =

B-6.

In how many ways we can select a comm ittee of 6 persons from 6 boys and 3 girls, if atleast two boys

65 = 15 1 2

& atleast two girls must be there in the com mittee? 6 yM+dksvkSj3yM+fd;ksaesals6O;fDr;ksadh,dlfefrfdrusrjhdksalscukbZtkldrhgS;fnlfefresadels

de nks yM+ds vkSj de ls de nks yM+fd;ksa dks vo'; 'kkfey fd;k tk;sA Sol.

Ans. 65 case-I If select 4 boys and 2 girls




Sol.

6C

4 ×

3C

2 =

15 × 3 = 45 case-II If select 3 boys and 3 girls 6C × 3C = 20 × 1 = 20  4 3 Hence total number of ways = 45 + 20 = 65 fLFkfr-I ;fn 4 yM+dso 2 yM+fd;ka pquh tk;s 6C × 3C = 15 × 3 = 45  4 2 fLFkfr-II ;fn 3 yM+dso 3 yM+fd;ka pquh tk;s 6C × 3C = 20 × 1 = 20  4 3 vr% dqy rjhdksa dh la[;k = 45 + 20 = 65

B-7.

In how many ways 11 players can be select ed from 15 players , if only 6 of these players can bowl and the 11 players must include atleast 4 bowlers? 15f[kykfM+;ksaesals11f[kykfM+;ksadh,dVhefdrusrjhdksalspquhtkldrhgS;fnbuf[kykfM+;ksaesalsdsoy 6 f[kykM+hxsanckthdjldrsgSa vkSj11 f[kykfM+;ksaesadelsde 4xsancktvo';'kkfeyfd;stk;saA Ans. 1170 Sol. Total No. of bowlers = 6 Now, (i) If 4 bowlers are including the no. of ways selecting 11 players out of 15 players = 6C4 × 9C 7 = 15 × 36 = 540 (ii) If 5 bowlers are selected = 6C 5 × 9C6 = 6 × 84 = 504 (iii) If all 6 bowlers are selected = 6C6 × 9C5 = 1 × 126 = 126 Hence total no. of ways = 540 + 504 + 126 = 1170 Hindi. dqy xsanckt = 6 (i) ;fn4xsanckt'kkfeyfd;stk;srks15f[kykfM+;ksaesals11 f[kykfM+;ksadsp;udsrjhds = 6C4 × 9C 7 = 15 × 36 = 540 (ii) ;fn5 xsanckt p;fur fd;s tkrs gSaA = 6C 5 × 9C6 = 6 × 84 = 504 (iii) ;fn 6 xsanckt p;fur fd;s tkrs gSaA = 6C6 × 9C5 = 1 × 126 = 126 vr% dqy rjhds = 540 + 504 + 126 = 1170 B-8.

In a question paper there are two parts part A and part B each consisting of 5 questions. In how many ways a student can answer 6 questions, by selecting atleast two from each part? ,diz'ui=kesanksHkkxgSHkkx AvkSjHkkxBrFkkizR;sdHkkxesa5iz'ugSaA,dNk=k6iz'uksadsmÙkjfdrusrjhdksa

ls ns ldrk gS ;fn mls izR;sd Hkkx esa ls de ls de nks iz'u vo'; p;u djus gSA Sol.

Ans. 200 Total no. of selected qu estion Part A Part B 2 4 3 3 4 2

Number of ways C2 × 5C 4 = 10 × 5 = 50 5 C3 × 5C 3 = 10 × 10 = 100 5 C4 × 5C 2 = 5 × 10 = 50 5

Total no. of ways = 200 Hindi.

dqyp;furiz'uksadhla[;k Hkkx A HkkxB 2 3 4

4 3 2

rjhdksadhla[;k 5

C2 × 5C 4 = 10 × 5 = 50 5 C3 × 5C 3 = 10 × 10 = 100 5 C4 × 5C 2 = 5 × 10 = 50

dqy rjhds = 200 B-9.

How many four digit natural numbers not exceeding the num ber 4321 can be formed using the digits 1, PERMUTATION & COMBINATION - 6

2, 3, 4, if repetition is allowed? vad 1,2,3,4 dh lgk;rkls pkj vadksa dh fdruh izkd`r la[;k,¡ cukbZ tk ldrh gStks4321 ls cM+h ugha gks] tcfd

vadksdhiqujko`fÙkgksldrhgSaA Sol.

Ans. 229 Total no. o f ways = 64

= 64

= 64

= 16

= 16

= 4 = 1 Adding = 229 Hindi.

adqyrjhds = 64

= 64

= 64

= 16

= 16

= 4 = 1

atksM+usij=229 B-10.

A committee of 6 is t o be chosen fro m 10 persons wi th the c ondition that if a parti cula r person 'A' is chosen, then another particular person B must be chosen. 10 O;fDr;ksaesals 6dh,dlfefrfdrusrjhdksalscukbZtkldrh gS;fn,dfo'ks"kO;fDr 'A' dkspquktkrkgS] rksnwljsfo'ks"kO;fDr'B'dksvo';pquukgksxk\ Ans. 154


If ‘A’ is not chosen, then number of selections = 9C6 If ‘A’ is chosen, then ‘B’ is also chosen, then nu mber of selection = 8C4  number of ways = 9C 6 + 8C 4 = 154 Hindi. ;fn ‘A’ dk p;u u gks] rks p;uksa dh la[;k = 9C6 ;fn ‘A’dk p;u gks] rks ‘B’ dk Hkh p;u djuk gS] rks p;uksa dh la[;k= 8C4  dqy rjhds = 9C6 + 8C4 = 154 Sol.

B-11.

In how many ways a team of 5 can be chosen from 4 girls & 7 boys, if the team has atleast 3 girls. 4 yM+fd;ksa vkSj 7 yM+dks esa ls 5dh,dVhefdrusrjhdksalspquhtkldrhgS ;fnVheesadelsde3yM+fd;k¡

vo';gks\ Sol.

Sol.

B-12.

Ans. (i)

91 If 3 girls and 2 boys are selected = 4C3 × 7C2 = 4 × 21 = 84 (ii) If 4 girls and 1 boys ar e selected = 4C4 × 7C1 = 1 × 7 = 7 Hence total number of ways = 84 + 7 = 91 ;fn 3 yM+fd;ksa o 2 yM+dksa dk p;u fd;k tk;s = 4C3 × 7C2 = 4 × 21 = 84 (i) (ii) ;fn 4 yM+fd;ksa o 1 yM+ds dk p;u fd;k tk;s = 4C4 × 7C1 = 1 × 7 = 7 vr% dqy rjhdksa dh la[;k = 84 + 7 = 91 In how many ways we can select 5 cards from a deck of 52 cards, if each selection must include atleast one king. 52 iÙkksadh,dxM~Mhesals 5iÙksfdrusrjhdksalspqustkldrsgSa;fn izR;sdp;uesadelsde,dckn'kkg

vo'; vk;s\ Sol.

Sol.

B-13.

Ans. 886656 Total number of ways = 4C 1 × 48C4 + 4C 2 × 48C3 + 4C3 × 48C 2 + 4C4 × 48C1 = 4 × 194580 + 6 × 17276 + 4× 1128 = 778320 + 103776 + 4512 + 1 × 48 = 886656 dqy rjhdksa dh la[;k = 4C1 × 48C4 + 4C2 × 48C3 + 4C3 × 48C2 + 4C4 × 48C1 = 4 × 194580 + 6 × 17276 + 4× 1128 = 778320 + 103776 + 4512 + 1 × 48 = 886656 Find number of ways of selection of atleast one vowel and one consonant from the word TRIPLE TRIPLE 'kCn ds v{kjksa esa ls de ls de ,d Loj rFkk ,d O;atu dk p;u fdrus izdkj ls dj ldrs gS\

Sol.

Ans. 45 Total number of ways out of 4 consonant and 2 vowels is 2C × { 4C + 4C + 4C + 4C } + 2C × {4C + 4C + 4C + 4C } 1 1 2 3 4 2 1 2 3 4 = 2 × {4 + 6 + 4 + 1} + 1 × [4 + 6 + 4 + 1] = 2 × 15 +15 = 45

Section (C) : Problem base on distinct and identical objects/devisors (PCTS/PCDV)

[k.M(C):nhxbZoLrqvksadsle:irFkkfHkUu&fHkUugksuslslEcfU/krleL;k C-1_.

Let N = 24500, then find (i) The number of ways by which N can be resolved into two factors. (ii) The number of ways by which 5N can be resolved into two factors. (iii) The number of ways by which N can be resolved into two coprime factors.

ekuk N = 24500 rks Kkr djks: (i) murjhdksadhla[;k]ftuesaNdksnksxq.ku[k.Mksaesafo;ksftrfd;ktkldrkgS (ii) murjhdksadhla[;k]ftuesa5Ndksnksxq.ku[k.Mksaesafo;ksftrfd;ktk ldrkgS PERMUTATION & COMBINATION - 8

Sol.

(iii)

murjhdksadhla[;k]ftuesaN dksnkslg&vHkkT;xq.ku[k.Mksaesafo;ksftrfd;ktkldrkgS

Ans. (i)

(i) 18 (ii) 2 3 2 N = 2 . 5 . 7

3.4.3 2 (ii)

23

(iii)

4

 18

5N = 2 2 . 5 4 . 7 2

3.5.3  1  23 2 (iii)

C-2.

N = 2 2 . 5 3 . 7 2 23 – 1 = 4

Find number of ways in which one or more letter be selected from the letters AAAABBCCCDEF.

v{kjksa AAAABBCCCDEF esa ls ,d ;k vf/kd v{kjksa dk p;u fdrus izdkj ls fd;k tk ldrk gS\ Sol.

Ans. 479. Total number of ways of selecting = 5 × 3 × 4 × 2 3 = 480. But in the above count also includes that way in which no letter is selected, Since we are supposed to find the number of ways in which at least one letter is selected, therefore the required numbers of ways = 480 – 1 = 479.

Hindi.

p;u ds dqy rjhds = 5 × 3 × 4 × 2 3 = 480. ijUrqmijksDrrjhdksaesaogrjhdkHkh'kkfeygSftlesadksbZHkhv{kjp;uesaughavk;kgSA pw¡fdgesa os rjhds Kkr djus gS ftlesa dels de ,d v{kj p;fur gqvk gSA vr% dqy vHkh"V rjhds = 480 – 1 = 479.

C-3.

Find number of divisiors of 1980. (i) How many of them are multiple of 11 ? find their sum (ii) How many of them are divisible by 4 but not by 15. 1980 ds Hkktdksa dh la[;k Kkr dhft,] muesa ls fdrus 11 ds xq.kt gS\ (i) (ii)

buesalsfdrus 11 dsxq.ktgS]budk;ksxHkhKkrdhft,A buesalsfdrus 4 lsHkkT;ysfdu15 lsughaA

Ans.

Sol.

Sol.

36, (i) 18, 11.(20 + 2 1 + 22) (3 0 + 3 + 3 2) (5° + 5) (ii) 3.2 + 1.1.2 = 8 2 2 1980 = 2 . 3 . 5 . 11, number of divisiors of 1980 = 36 (i) 3.3.2 = 18 sum = 11.(1 + 2 + 2 2) . (1 + 3 + 3 2) . (1 + 5) (ii) 3.2 + 1.1.2 = 8 1980 = 2 2 . 3 2 . 5 . 11, 1980 ds Hkktdksa dh la[;k = 36 (i) 3.3.2 = 18 Hkktdksa dk ;ksx = 11.(1 + 2 + 22) . (1 + 3 + 3 2) . (1 + 5) (ii) 3.2 + 1.1.2 = 8

Section (D) : Problem based on circular arrangement / Multinomial theorem (PCMT/PCCA)

[k.M (D):o`Ùkh;Øep;rFkkcgqinh;izes;ijvk/kkfjrleL;k,¡ PERMUTATION & COMBINATION - 9

D-1.

There are 3 white, 4 blue and 1 red flowers. All of them are tak en out one by one and arranged in a row in the order. How many different arrangements are possible (flowers of sam e colurs are similar)? 3 lQsn] 4 uhys vkSj 1 yky Qwy esa ls ,d ds ckn ,d Qwy dks fudkydj mUgsa ,d iafDr esa O;ofLFkr fd;k tkrk

gSaA,slhfdruhfofHkUuO;oLFkk,¡lEHkogSa\¼lekujaxdsQwylekugSaA½ Sol.

Ans. 280 Total number of possible arrange ments are

8! = 280 3! 4! 8!

dqy lEHko foU;klksa dh la[;k = 3! 4! = 280 D-2.

In how many ways 5 persons can sit at a round table, if two of the persons do not sit to gether? ,dxksyestdspkjksavksj5O;fDrfdrusrjhdksalscSBldrs gSa;fnbuesalsnksfo'ks"kO;fDr ,dlkFkughacSBrs

gSaA Sol.

Ans. 12 First find if all the person are sitting in a round table is 4! = 24 ways if two of the person are sitting together i.e. 3! × 2! = 12 ways Hence required number of ways = 24 – 12 = 12 ways

Sol.

;fnlHkhO;fDrxksyestdspkjksarjQcSBrsgSa]rksrjhdksadhdqyla[;k4!=24 rjhds ;fnnksfo'ks"kO;fDr,dlkFkcSBrsgSa]rksrjhdksadhla[;k =3!×2!=12 rjhds vr% vHkh"V rjhdksa dh la[;k = 24 – 12 = 12 rjhds

D-3.

In how many ways four men and three women may sit around a round table if all the women are together?

pkjiq:"krFkkrhuefgyk,sa,dxksyest dspkjksavkSjfdrusizdkjls cSBldrsgSa]tcfdlHkhefgyk,salkFkesa cSBsA Sol.

Ans. 144 Total number of women are sit together then total number of person is 5 hence required ways = 4! × 3! = 24 × 6 = 144

Sol.

lHkhefgykvksadks,dlkFkcSBkrsgq,O;fDr;ksadksxksyestdhpkjksvkSjcSBusdsdqyrjhds=4!×3! = 24 × 6 = 144

D-4.

Seven persons including A, B, C are s eated on a circular table. How m any arrangements are possible if B is always between A and C? A,B, C lfgr lkr O;fDr ,d xksy eSt ds pkjksa vkSj fdrus izdkj ls cSB ldrs gSa tcfd B lnSo ArFkkC ds

e/;cSBsA Sol.

Ans. 48 Here B is always between A and C so i.e. either ABC or CBA so total required number of ways is 4! × 2! = 24 × 2 = 48 ways

Sol.

;fn B lnSo Ao C ds lkFk cSBrk gS rks ;k rks ABC ;k CBA vr% dqy vHkh"V rjhdksa dh la[;k 4! × 2! = 24 × 2 = 48 rjhds

D-5_.

In how many ways four '+' and five '–' sign can be arranged in a circles so that no two '+' sign are together.

pkj'+'rFkkikap'–' fpUg~ksadks,do`ÙkdspkjksavkSjfdrusizdkjlsO;ofLFkrfd;ktkldrkgS;fndksbZHkhnks '+'fpUg~lkFkugksaA Sol. Sol.

1 Ans. By simple diagram it is obvious that there is only one way.

ljy fp=k dh lgk;rk ls ;g Li"V gS fd ,slk ,d gh rjhdk lEHko gSA


D-6.

In how many ways fifteen different items may be given to A, B, C such that A gets 3, B gets 5 and remaining goes to C. 15fofHkUuoLrqvksadks A, BrFkkC esafdrusizdkjlsck¡Vktkldrk gS]tcfd Adks 3, B dks 5rFkk'ks"koLrq,sa CdksnsrsgksA

Sol.

Ans. 360360 No. of ways 3 item can be given to A is 15 C3 then no. of ways 5 item can be given to B is 12 C5 Rest are given to C is 7C7 Hence total no. of ways if all item is given is 15 C3 × 12 C 5 × 7C 7 = 455 × 792 × 1 = 360360

Hindi. A dks 3 oLrq,¡ nsus ds dqy rjhds = 15C3

rc B dks 5 oLrq,¡nsus dqy rjhds = 12C5 vr%'ks"kcphgqbZoLrq,¡C dks7C7rjhdslsnhtkldrhgSA ;fnlHkhoLrq,¡nhx;hgks]rksdqyrjhds 15

D-7.

C3 × 12 C 5 × 7C 7

= 455 × 792 × 1 = 360360

Find number of ways of distributing 8 different items equally among two children. 8 fofHkUuoLrqvksadksnkscPpksaesaleku:ilsfdrusizdkjlsck¡VldrsgSA

Sol.

Hindi.

Ans. 70 Total no. of ways = First distribute 4 items from 8 to a child, then remaining 4 item to other = 8C 4 × 4 C 4 = 70 × 1 = 70

dqyrjhds=igys4oLrqvksadks8esals,dcPpsdksnsrsgaS]bldsi'pkr~cphgqbZ 4oLrq,¡nwljscPPksdksnsrsgSaA = 8C 4 × 4 C 4

D-8.

= 70 × 1 = 70

Find number of negative integral solution of equation x + y + z = – 12

lehdj.k x + y + z = – 12 ds _.kkRed iw.kk±d gyksa dh la[;k Kkr dhft,A Sol.

Ans. 55 Here –10  x, y, z  – 1 Using multinomial theorem Find the coefficient of x 12 in this expansion of (x + x 2 + .......+ x 10) 3 = x 3 (1 + x + x 2 + .......+ x 9)3 = x3 (1 – x 10 )3 . (1 – x) –3

11 10 = 55 2 Hindi. ;gk¡ –10  x, y, z  – 1 =

11

C 9 =

cgqinh;xq.kuizes;dsiz;ksxls (x + x2 + .......+ x10)3 ds izlkj esa x12 dk xq.kkad Kkr djus ij = x 3 (1 + x + x 2 + .......+ x 9)3 = x3 (1 – x 10 )3 . (1 – x) –3

11 10 = 55 2 In how many ways it is possible to divide six identical green, six identical blue and six identical red among two persons such that each gets equal number of item? =

D-9.

11

C 9 =

N%,dlekugjh]N%,dlekuuhyh rFkkN%,dlekuykyoLrqvksadks nksO;fDr;ksaesafdrusizdkjlsck¡Vldrs gSa]tcfdizR;sdO;fDrlekula[;kesaoLrq,saizkIrdjrkgksA Sol.

Ans. 37 x 1 + x 2 + x 3 = 9 Coefficient of x 9 in (1 + x + ...... + x 6)3 PERMUTATION & COMBINATION - 11

Sol.

Coefficient of x 9 in (1 – x 7) 3 (1 – x) –3 Coefficient of x 9 in (1 – 3x 7) (1 – x) –3 = 9+3–1C 2 – 3 . 2 + 3 – 1 C 2 = 55 – 18 = 37 x 1 + x 2 + x 3 = 9 ds izlkj esa x9 dk xq.kkad (1 + x + ...... + x6)3

ds izlkj esa x9 dk xq.kkad (1 – x7)3 (1 – x) –3 ds izlkj esa x9 dk xq.kkad (1 – 3x7) (1 – x) –3 = 9+3–1C2 – 3 . 2 + 3 – 1C2 = 55 – 18 = 37 D-10.

Find total number of positive integral solutions of 15 < x 1 + x 2 + x 3  20. 15 < x1 + x2 + x3  20ds dqy /kukRed iw.kk±d gyksa dh la[;k Kkr dhft,A

Sol.

Ans. 685 x 1 + x 2 + x 3 = 20 – t t = 0, 1, 2, 3, 4 4

Required value =

Sol.



19 – t

C2

t 0

= 20C3 – 15C 3 = 1140 – 455 = 685 x 1 + x 2 + x 3 = 20 – t t = 0, 1, 2, 3, 4 4

vHkh"V eku

=



19 – t

t 0

C2 = 20C – 15C = 1140 – 455 = 685 3 3

Section (E) : Problem based on geometry / Dearrangement / exponent of prime/ Principal of exculusion/Grouping (PCGT/PCDA/PCGP)

[k.M (E) :T;kfefr ij vk/kkfjr] vO;oLFkk ij] vHkkT; la[;k dh ?kkr] foHkktu dk fl)[email protected] E-1_.

In how many ways 18 diffrent objects can be divided into 7groups such that four groups contains 3 objects each and three groups contains 2 objects each. 18fofHkUuoLrqvksadks7lewgksaesafdrusizdkjlsfoHkkftrfd;ktk ldrkgS;fnbuesals 4 lewgksaesalsizR;sdesa3,oa'ks"k3lewgksaesalsizR;sdesa2oLrq,¡gksa\ Ans.

Sol.

E-2.

Sol.

18! 4

(3! ) . (2! )3 4! 2! 18!

(3! )4 . (2! )3 4! 2!

(a)

In how many ways can five people be divided into three groups?

(b) (c)

In how many ways can five people be distributed in three different rooms if no room must be empty? In how many ways can five people be arranged in three different rooms if no room must be empty?

(a)

5O;fDr;ksasdks3lewgksaesafdrusizdkjlsfoHkkftrfd;ktkldrkgS\

(b)

5O;fDr;ksadks3fofHkUudejksaesafdrusizdkjlsck¡VktkldrkgS;fndksbZHkhdejk[kkyhujgs\

(c)

5O;fDr;ksadks3dejksaesafdrusizdkjlsO;ofLFkrfd;ktkldrkgS;fndksbZHkhdejk[kkyhujgs\

Ans. (a)

(a)

25

(b)

150.

(c)

720

five people can be divided into three groups in the following way ; 1, 1, 3 or 1, 2, 2 Hence, total number of ways


=

1 1 5! 5! × + × 2! 2! 3! ( 2 ! )2

= 10 + 15 = 25 (b)

The three different rooms can be filled in the following ways;

  

Room

People 1 1

Room or

3

  

People 1 2

2

Number of ways = ways in which such groups can be formed ways in which the groups can be × arranged in the three different rooms Case -

Number of ways =

 5 ! 1     3 ! = 60  3 ! 2 ! 

Case -

Number of ways =

 5 ! 1      (2 ! )2 2 !  3 ! = 90  

Hence, required number of ways = 60 + 90 = 150. (c) In this case, the positioning of the people amongst themselves is also to be taken into account. Case Number of ways = 60 × 3 ! = 360 Case Number of ways = 90 × 2 ! × 2 ! = 360 Hence, required number of ways = 360 + 360 = 720. Hindi. (a)

5O;fDr;ksadks3lewgksaesafuEurjhdkslsfoHkkftrfd;ktkldrkgSA 1, 1, 3

;k

1, 2, 2

vr%dqyrjhds =

5! 5! 1 1 × + 2 × 3! 2! 2! (2 ! )

= 10 + 15 = 25 (b)

3 fofHkUudejksadksfuEuizdkjlsHkjktkldrkgSA

Room

  

People 1 1

Room or

3

  

People 1 2

2

dqyrjhds=lewgksadkscukusdsrjhds×lewgksadksrhudejksaesaO;ofLFkrdjusds rjhds  5 ! 

1 

fLFkfr- dqy rjhds =  3 !  2 !  3 ! = 60 fLFkfr-



 5 ! 1    dqy rjhds =  (2 ! )2 2 !  3 ! = 90  

vr% vHkh"V rjhds = 60 + 90 = 150. (c) blfLFkfresaO;fDr;ksadksvkilesafLFkfrdksHkhx.kukesaysukgksxkA fLFkfrdqy rjhds = 60 × 3 ! = 360 fLFkfrdqy rjhds = 90 × 2 ! × 2 ! = 360 vr% vHkh"V rjhds = 360 + 360 = 720.


E-3_.

200!

Prove that :

(10! )20 19! 200!

fl) dhft, : Sol.

200! (10! )20 .20! 200!

=

(10! )20 .19!

× 20

which must be an integer..

200oLrqvksadks20lewgksaesa]izR;sdesa10oLrq,agS]ckaVusdsdqyrjhds=

=

E-4.

,d iw.kk±d gSaA

Number of ways of distributing 200 objects into 20 groups each containing 10 objects =

Sol.

(10! )20 19!

is an integer

200! (10! )20 .19!

200! (10! )20 .20!

×20

tksfd,diw.kkZadla[;kgks

Find exponent of 3 in 20 ! 20 ! esa 3 dk ?kkrkad Kkr dhft,A

Sol.

Ans. 8 Exponent of 3 in 20!

 Sol.

 20   20   20   3    32    35  + ........

=6+2+0=8 20!aesa3dk?kkrkad



 20   20   20   3    32    35  + ........ =6+2+0=8

E-5.

Find number of zeros at the end of 45! . 45! dsvUresa'kwU;ksadhla[;kKkrdhft,A

Sol.

Ans. 10 Exponent of 2 in 45! is

 45   45   45   45   45   45   2  +  2 2  +  2 3  +  2 4  +  2 5  +  2 6              = 22 + 11 + 5 + 2 + 1 + 0 Exponent of 5 in 45! is

= 41

 45   45   45   5  +  5 2  +  5 3  = 9 + 1 + 0 = 10       So no. of zeros at the end of 45! is 10 Hindi. 45! esa2dk?kkrkad

 45   45   45   45   45   45   2  +  2 2  +  2 3  +  2 4  +  2 5  +  2 6              = 22 + 11 + 5 + 2 + 1 + 0

= 41

45! esa5dk?kkarkd PERMUTATION & COMBINATION - 14

 45   5 +  

 45   5 2  +  

 45   5 3  = 9 + 1 + 0 = 10  

vr%45! dsvaresa'kwU;ksadhla[;k10gksxhA E-6.

A pe rson writes letters to f ive f riends and addr esses on the c orr espondin g en velopes. In how many ways can the letters be placed in the envelopes so that (a) all letters are in the wrong envelopes? (b) at least three of them are in the wrong envelopes?

,dO;fDrvius5 fe=kksadksi=kfy[krkgSvkSjlaxrfyQkQksijirsfy[krkgSA;si=kfyQkQksaesafdrusizdkjls j[kstkldrsgSarkfd (a) lHkhi=kxyrfyQkQksaesagks\ (b) delsderhui=k xyrfyQkQksaesagksa\

Sol.

Ans.

(a)

44

(b)

(a)

Required number of ways = 5 ! 1 

(b)

Required num ber of ways = 5C 2 D3 + 5C1 D4 + 5C 0 D5

 

 

= 10 × 3 ! 1 

(b)

1 1 1 1 1       = 44 1 ! 2 ! 3 ! 4 ! 5 ! 

1 1 1     + 5 × 4 ! 1 ! 2 ! 3 ! 

+1×5!

Hindi. (a)

109

 1 1 1 1  1       1 ! 2 ! 3 ! 4 ! 

 1 1 1 1 1  1       = 109  1 ! 2 ! 3 ! 4 ! 5 !   

vHkh"V rjhdksa dh la[;k= 5 ! 1 

1 1 1 1 1       = 44 1 ! 2 ! 3 ! 4 ! 5 ! 

vHkh"V rjhdksa dh la[;k= 5C2 D3 + 5C1 D4 + 5C0 D5  

= 10 × 3 ! 1 

1 

   + 5 × 4 ! 1  1  1  1  1  1 ! 2 ! 3 !   1 ! 2 ! 3 ! 4 !  1

+1×5!



1



 1 1 1 1 1  1       = 109  1 ! 2 ! 3 ! 4 ! 5 ! 

PART - II : OBJECTIVE QUESTIONS

Hkkx- II : oLrqfu"B iz'u ¼OBJECTIVEQUESTIONS ½ * Marked Questions are having more than one correct option. * fpfUgr iz'u ,d ls vf/kd lgh fodYi okys iz'u gS -

Section (A) : Problem based on Problems, Arrangements of given objects/Selection of given object (PCAD/PCSD)

[k.M (A):nhxbZoLrqvksadspquko]foU;klijvk/kkfjrleL;k,¡


A-1.

Sol.

Hindi.

Let Pm stand for mPm. Then the expression 1. P 1 + 2. P2 + 3. P3 +...... + n. P n = (A*) (n + 1) !  1 (B) (n + 1) ! + 1 (C) (n + 1) ! (D) none m ;fn Pm dks Pm ls n'kkZ;k tkrk gks] rks O;atd 1. P1 + 2. P2 + 3. P3 +...... + n. Pn = (A) (n + 1) !  1 (B) (n + 1) ! + 1 (C) (n + 1) ! (D) buesalsdksbZugha Here T r = r. Pr = (r + 1–1) r ! = (r + 1) r! – r! = (r + 1)! – r!  S n = (n+1)! –1

;gk¡ Tr = r. Pr = (r + 1–1) r ! = (r + 1) r! – r! = (r + 1)! – r!  Sn = (n + 1)! –1

A-2.

Sol. Sol.

The number of signals that can be made with 3 flags each of different colour by hoisting 1 or 2 or 3 above the other, is: fofHkUu jaxks ds 3 >.Mksa dk mi;ksx djds dqy fdrus ladsr cuk, tk ldrs gSa tcfd ,d >.Ms ds Åij1 ;k 2 ;k 3 >.Ms Qgjk;s tk ldrs gSa\ (A) 3 (B) 7 (C*) 15 (D) 16 Total number of signals can be m ade from 3 flags each of different colour by hoisting 1 or 2 or 3 above. 3p + 3p + 3p = 3 + 6 + 6 = 15 i.e. 1 2 3 fofHkUujaxksads3 >.Mksadksmi;ksxdjcuk;sx;sdqyladsrksadhla[;k]tcfd,d>.MsdsÅij 1;k2;k3 >.Ms

Qgjk;stkldrsgSaA i.e. A-3.

3p

1 +

3p

2 +

3p

3 =

3 + 6 + 6 = 15

10 different letters of an alphabet are given. W ords with 5 letters are formed fr om these given letters, then the number of words which have atleast one letter repeated is: (A*) 69760 (B) 30240 (C) 99748 (D) none vaxzstho.kZekykds10 v{kjfn;sx;sgSaAbuv{kjksadhlgk;rkls 5v{kjokys'kCncuk;stkrsgS]rks,slsfdrus

'kCn gksxsaftuesadelsde,dv{kjdhiqujko`fÙkgksrhgS\

(A) 69760 (B) 30240 (C) 99748 (D)buesalsdksbZugha Sol. Number of words which have at least one letter repeated = total words – number of words which have no letter repeated = 10 5 – 10 × 9 × 8 × 7 × 6 = 69760 Hindi. ‘'kCnksadhla[;kftuesadels de,dv{kjdh iqujko`fÙkgks= dqy'kCn – mu'kCnksadhla[;k ftuesa fdlh v{kj dh iqujko`fÙk uk gks= 105 – 10 × 9 × 8 × 7 × 6 = 69760 A-4.

Sol.

The number of numbers from 1000 to 9999 (both inclusive) that do not have all 4 different digits, is: 1000ls9999 dschp(nksuksa'kkfeygSa),slhfdruhla[;k,¡gksaxhftudspkjksavadvyx&vyxughagSa\ (A) 4048 (B*) 4464 (C) 4518 (D) 4536 There are four digit no which is atleast one is repeated i.e. (do not all are different) is – = 9000 – 4536 = 4464

Sol.

;fnpkj vadksa dh og la[;k,a Kkr djuh gS ftuesade ls de,d vaddh iqujko`fÙk gks jgh gS¼lHkhvad vyx&vyx ughagS½ – = 9000 – 4536 = 4464

Section (B) : Problem based on Selection as well as arrangement of objects/Rank of word (PCSR/PCSI)

[k.M (B):nhxbZoLrqvksadspquko],oamudsfoU;klijvk/kkfjrleL;k,¡]fn;sx;s'kCndhojh;rkij vk/kkfjrleL;k,¡


B-1.

8 chairs are numbered from 1 to 8. Two women & 3 men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4, then the men select the chairs from among the remaining. The number of possible arrangements is:

vkBdqflZ;ksaij1 ls8rduEcjyxsgSaSA 2vkSjrsa vkSj3 vknehizR;sd,d&,ddqlhZijcSBukpkgrsgSaAigys vkSjrsa1ls4 uEcjyxhgqbZdqflZ;ksaesalsdqflZ;k¡pqurhgSavkSjbldscknvknehcphgqbZdqflZ;ksaesalsdqflZ;k¡pqurs gS]rkslaHkkforfoU;klksadhla[;kgksxh\ Sol. Sol.

(A) 6C3. 4C4 (B) P2. 4P3 Total number of possible arran gements is 4 p2 × 6p3 .

(D*) 4P2. 6P3

dqylaHkofoU;kvksadhla[;k 4

B-2.

(C) 4C3. 4P3

p2 × 6p3.

Number of words that can be made with the letters of the word "GENIUS" if each word neither begins with G nor ends in S, is: "GENIUS" 'kCndsv{kjksals,slsfdrus'kCncuk,tkldrsgSatksurks G ls'kq:gksrsgSavkSjugh S ijlekIr

gksrsgSa\ Sol.

Sol.

(A) 24 (B) 240 (C) 480 (D*) 504 First we have to find all the arrangements of the word ‘GENIUS’ is 6 ! = 720 number of arrangement which in either started with G ends with S is (5! + 5! – 4! ) = (120 + 120 – 24)= 216 Hence total number of arrangement which is neither started with G nor ends with S is. (720 – 216) = 504

loZizFkege'kCn‘GENIUS’ dslHkhlaHkofoU;klksadhla[;kKkrdjsxsaA 6 ! = 720

mufoU;klksadhla[;ktks;krks GlsizkjEHkgksrksSijlekIrgks (5! + 5! – 4! ) = (120 + 120 – 24)= 216 vr%mufoU;klksadhla[;ktksurks G lsizkjEHkgksrsgSavkSjughSijlkeIrgksrsgSa (720 – 216) = 504 B-3.

5 boys & 3 girls are sitting in a row of 8 seats. Number of ways in which they can be seated so that not all the girls sit side by side, is: 5 yM+dsvkSj3yM+fd;k¡,diafDresafLFkr8lhVksaijcSBsgq,gSaA;fnlHkhyM+fd;k¡,dlkFkughacSBs]rksoslHkh

fdrusrjhdksalscSBldrsgSa\ Sol.

Hindi.

(A*) 36000 (B) 9080 (C) 3960 Total no. of arrangem ent if all the girls do not seat side by side is = [all arrangement – girls seat side by side] = 8! – (6! × 3!) = 6! (56 –6) = 6! × 50 = 720 × 50 = 36000

(D) 11600

;fnlHkhyM+fd;k¡,dlkFkughacsBsarksdqyfoU;klksadhla[;k = [dqyfoU;kl – tclHkhyM+fd;kalkFk&lkFkcSBs] = 8! – (6! × 3!) = 6! (56 –6) = 6! × 50 = 720 × 50 = 36000

B-4.

The sum of all the numbers which can be form ed by using the digits 1, 3, 5, 7 all at a tim e and which have no digit repeated, is

vadks 1,3,5,7 dk,dlkFkmi;ksxdjrsgq,cukbZtkldusokyhlHkh la[;kvksadk;ksxfdrukgksxktcfdvadks


dhiqujko`fÙkughagksldrhgSa\ Sol.

Hindi.

(A) 16 × 4! (B) 1111 × 3! (C*) 16 × 1111 × 3! If 1 be unit digit then total no. of number is 3! = 6 Similarly so on if 3, 5, or 7 b e unit digit number then total no. of no. is 3! = 6 Hence sum of a ll unit digit no. is = 6× (1+3+5+7) = 6× 16 = 96 Hence total sum is = 96 × 10 3 + 96 × 10 2 + 96 × 10 1 + 96 × 10 0 = 96000 + 9600 + 960 + 96 = 106656 = 16 × 1111 × 3!

(D) 16 × 1111 × 4!.

;fnbdkbZ dk vad 1 gks] rks dqy la[;kvksa dh la[;k = 3! = 6 blh izdkj ;fn 3, 5 ;k7 bdkbZ vad gks] rks dqy la[;kvksa dh la[;k = 3! = 6 vr% lHkh bdkbZ vadksa dh la[;kvksa dk ;ksx= 6× (1+3+5+7) = 6× 16 = 96 vr% dqy ;ksx = 96 × 10 3 + 96 × 10 2 + 96 × 10 1 + 96 × 10 0 = 96000 + 9600 + 960 + 96 = 106656 = 16 × 1111 × 3!

B-5.

The number of words that can be for med by using the letters of the wor d ‘MATHEMATICS’ that start as well as end with T, is ‘MATHEMATICS’ 'kCndsv{kjksals,slsfdrus'kCncuk,tkldrsgSa tksT lsizkjEHkgksrsgSavkSj T ijghlekIr

gksrsgS\ (A) 80720

(B*) 90720

(C) 20860

(D) 37528

Sol. Total arrangem ent is

9! 2!. 2!

= 90720 Sol.

9!

dqy foU;klksa dh la[;k = 2!. 2! =90720 B-6.

How many nine digit numbers can be formed using the digits 2, 2, 3, 3, 5, 5, 8, 8, 8 so that the odd digits occupy even positions?

vadks 2, 2, 3, 3, 5, 5, 8, 8, 8 dh lgk;rk ls 9 vadks dh fdruh la[;k,¡ cukbZ tk ldrh gSa] tcfd fo"ke vad le LFkkuksaijvk;s? Sol.

(A) 7560 Even place

(B) 180

(C) 16

(D*) 60

There are four even places and f our odd digit number so total num ber of filling is

occupy is

2!. 2!

rest are also

5! ways 3!. 2!

Hence total number of ways = Sol.

4!

4! 2!. 2!



5! 3!. 2!

= 60

leLFkku PERMUTATION & COMBINATION - 18

4!

;gkapkjleLFkkugSopkj fo"kevadgSAvr%fo"kevadksadksleLFkkuksaijHkjusdsrjhds 2!. 2! gSrFkkvU;vadksa 5!

dks 3!. 2! rjhdksalsHkjktkldrkgSA 4!

5!

vr % dqy rjhdksa dh la[;k = 2!. 2!  3!. 2! = 60 B-7.

The number of permutations that can be formed by arranging all the letters of the word ‘NINETEEN’ in which no two E’s occur together. is

'kCn‘NINETEEN’ dslHkhv{kjksadksfdrusizdkjls O;ofLFkrfd;ktkldrkgSarkfddksbZHkh nks'E',dlkFkugha vk;s\ 8! (A) 3! 3! Sol.

5! (B)

6

3!  C 2

(C*)

5! 6 3 ! × C3

(D)

8! 6 5 ! × C3.

NINETEEN  N  3 : , T E  3 First we arrange the word of N, N, N, I and T

5! . 3! Now total 6 number of place which are arrange E is 6C3 then the number of ways =

Hence total number of ways = Sol.

5! 6 . C3 3!

NINETEEN  N  3 : , T E  3 loZizFkege N,N,N,I vkSjT dks O;ofLFkr djsxsaA

vr% rjhdksa dh la[;k =

5! . 3!

vc 6 LFkkuksa ij E-3 dks j[kus ds rjhds =6C3 vr% dqy rjhdksa dh la[;k = B-8.

5! 6 . C3 3!

Out of seven consonants & four vowels, the number of words of six letters, formed by taking four consonants & two vowels, is (Assume each ordered group of letter is a word) 7 O;atuksarFkk4Lojksaesals 4 O;atu rFkk2 LojysdjN%v{kjksaokysfdrus'kCncuk;stkldrsgS\& (ekukv{kjksadkizR;sdØferlewg,d'kCngS)

Soll

(A) 210 (B) 462 (C*) 151200 (D) 332640 First we select 4 consonant out of 7 consonant and 2 vowel out of 4 vowel then arranging Hence total number of ways = 7C4 × 4C2 × 6!


Soll

= 210 × 720 = 151200 loZizFkege7O;atuksaesals4O;atupqusaxsarFkk4Lojksaesals3LojpqusaxsarFkkO;ofLFkrdjsxsaA

vr% dqy rjhdksa dh la[;k = 7C4 × 4C2 × 6! = 210 × 720 = 151200 B-9.

A box contains 2 wh ite b alls, 3 black balls & 4 r ed balls. In h ow many ways can three ba lls b e dra wn from the box if atleast one black ball is to be included in draw (the balls of the same colour are different). (A) 60 (B*) 64 (C) 56 (D) none

,dlUnwdesa2fHkUulQsn]3fHkUudkyhvkSj4 fHkUuykyxsnsagSaAlUnwdesals3xsansadqyfdrusrjhdksalsfudkyh tk ldrh gS tcfd izR;sd p;u esa de ls de ,d dkyh xsan vo'; lfEefyr gks\ (A) 60 (B) 64 (C) 56 (D)buesalsdksbZugha Sol.

Sol.

First we find 3 ball from 9 ball 9C = 84 3 Now number of ways if any one black ball not selected is 6C 3 = 20 Here required no is 84 - 20 = 64 loZizFke9xsnksaesals3xsanspquusdsrjhds=9C3=84

vr%murjhdksadhla[;kftuesadksbZHkhdkyhxsanughapquhxbZgks =6C3 =20 vr% vHkh"V rjhdksa dh la[;k 84 – 20 = 64 B-10.

Passengers are to travel by a double decked bus which can accommodate 13 in the upper deck and 7 in the lower deck. The number of ways that they can be divided if 5 refuse to sit in the upper deck and 8 refuse to sit in the lower deck, is

,d nks eaftyk cl ds Åijh eafty ij 13rFkk fupyh eafty ij 7 O;fDr cSB ldrs gSaA ;fn bu 20 O;fDr;ksa esa ls5 O;fDrÅijcSBuslseukdjrsgSarFkk 8 O;fDruhpsacSBuslseukdjrsgSa]rksbUgsfdrusizdkjlsfoHkkftr fd;k tk ldrk gSa\ Sol.

(A) 25 (B*) 21 (C) 18 upperdeck - 13 seats  8 in upper deck. lowerdeck - 7 seats  5 in lower deck Remains passengers = 7 Now Remains 5 seats in upper deck and 2 seats in lower deck for upper deck number of ways = 7C 5 for lower deck number of ways = 2C2

(D) 15

7 .6 = 21 2  Åijh eafty esa 8 ;k=kh

So total number of ways = 7C5 × 2C 2 = Hindi.

Åijh eafty - 13 lhVsa fupyh eafty- 7 lhVsa  fupyh eafty esa 5 ;k=kh cpsgq,;k=kh=7 vr% 5 lhVsÅijheaftyesarFkk 2 lhVsfupyheaftyesacprhgSaA Åijheaftydsfy;slhVsaHkjusdsrjhds =7C5 uhpsdheaftyesalhVsaHkjusdsrjhds= 2C2 vr% dqy rjhds = 7C5 × 2C2 =

7. 6 = 21 2

B-11*. In an examination, a candidate is required to pass in all the four sub jects he is studying. The number of ways in which he can fail is

,dijh{kkesa],dijh{kkFkhZdksmulHkhpkjfo"k;ksaesaiklgksukvko';dgSaftUgsogi<+jgkgSaAogfdrusrjhdksalsQsy gksldrkgSa\ (A) 4P1 + 4P2 + 4P3 + 4P4 (C*) 24 – 1

(B) 44 – 1 (D*) 4C1 + 4C2 + 4C3 + 4C 4


Sol. Sol.

Number of ways he can f ail is either one or t wo, three or four subject then total of ways. 4C + 4C + 4C +4C = 2 4 – 1 1 2 3 4

mldsvlQygksusdsrjhdksaesaog;krks,dfo"k;esa;knksfo"k;esa;krhufo"k;esa;kpkjfo"k;ksaesavlQygksldrk gks] vr% dqy rjhdksa dh la[;k =4C1 +4C2 + 4C3 +4C4 = 24 – 1

B-12*. In a cricket m atch against Pakistan, Azhar wants to bat before Jadeja and Jadeja wants to bat before Ganguli. Number of possible batting orders with the above restrictions, if the remaining eight team members are prepared to bat at any given place, is: (A*)

11 !

(B*) 11C3. 8 !

3!

(C)

11 !

(D) none

3

ikfdLrkudsf[kykQ,dfØdsVeSpesavtgj]tMstklsigyscSafVxdjukpkgrkgSvkSjtM+stk]xkaxqyhlsigyscSafVxdjuk pkgrkgSA;fn'ks"kvkBf[kykM+hfdlhHkhØeijcYysckthdsfy,rS;kjgks]rkslEHkkforcYysckthØeksadhla[;kgksxh – (A) Sol.

11 !

(B) 11C3. 8 !

3!

11 !

(D)buesaslsdksbZugha

3

First we choose any three place out of 11 place i.e. 11C3 ways and rest 8 places are arranged by 8! ways. Hence required no. is

Hindi.

(C)

11

C 3. 8! =

11 ! 3

igysge11LFkkuksaesalsdksbZ3LFkkuksadkp;udjsxsa vFkkZr~11C3 rFkk'ks"k 8LFkkuksadks8!fof/k;ksa}kjkO;ofLFkrfd;ktkrldrkgSA vr% okafNr la[;k 11C3. 8! =

11 ! 3

B-13*. The number of ways of arranging the letters AAAAA, BBB, CCC, D, EE & F in a row if the letter C are separated from one another is:

v{kjksa AAAAA,BBB,CCC,D,EEvkSjFdks,diafDresaO;ofLFkrdjusdsrjhdksdhla[;kKkrdjksgSa ;fnv{kjc,d nwljslsvyxjgs& (A*) 13C3. Sol.

12! 5 ! 3 ! 2!

(B)

13 ! 5 ! 3! 3 ! 2!

(C)

14 ! 3 ! 3! 2!

(D*) 11.

13 ! 6!

We have arrange all the letter except ‘ccc’ is

12! new there all 13 place where ‘i’ can be placed 5 !. 3 !. 2! Hence required number of ways is =

12! 5! 3! 2!

13 C

3 =

Hindi. ‘ccc’ dksNksMdjlHkhv{kjksadksO;ofLFkrdjusdsrjhds=

13 C

11 .

3

13! 6!

12! 5 !. 3 !. 2!

u;s13 LFkkutgk¡ ‘i’ dksO;ofLFkrfd;ktkrkgS]dsrjhds=13C3 12!

vr% vHkh"V rjhds = 5! 3! 2! 13C3 = 11 .

13! 6!

B-14*. The kindergarten teacher has 25 kids in her class. She takes 5 of them at a time, to zoological garden as often as she can, without taking the same 5 kids more than once. Then the number of visits, the teacher makes to the garden exceeds that of a k id by: PERMUTATION & COMBINATION - 21

,dulZjhd{kkdhf'kf{kdkdhd{kkesa25 cPpsgSaAogmudks5–5dslewgesafpfM+;k?kjfn[kkusystkrhgSa]tcfdmUgha 5 cPpksadslewgesalsfdlhdksHkh,dlsvf/kdckjuystk;ktkrkgks]rksf'kf{kdkfdlh,dcPpsdsHkze.kksadhla[;k lsfdruhT;knkckjHkze.kdjsxh& Sol.

(A*) 25C5  24C4 (B*) 24C5 (C) 25C5  24C5 Total no. of visits that a teacher goes is = 25C5 (selection of 5 different kids each time & teacher goes every time) Number of visits of a boy = select one particular boy & 4 from rest 24 = 24C4 So extra visits of a teacher from a boy is =

HINDI.

25

C5 – 24C4 =

24

(D) 24C4

C5

f'kf{kdkdsHkze.kksadhla[;k=25C5 (izR;sdckj5fHkUucPpsrFkkf'kf{kdkizR;sdckjesatkrhgS) ,d cPps ds Hkze.kksa dh la[;k =,d fof'k"B cPps dk p;u rFkk ckfd 24esa ls 4=24C4 vr% f'kf{kdk dk cPps ls vfrfjDr Hkze.k =25C5 –24C4 = 24C5


[k.M(C):nhxbZoLrqvksadsle:irFkkfHkUu&fHkUugksuslslEcfU/krleL;k C-1.

The number of divisors of a pbqc r ds where a, b, c, d are primes & p, q, r, s

 N, excluding 1 and the

number itself, is:

la[;kapbqcr ds ¼tgk¡a,b,c,d vHkkT;la[;k,¡gSavkSjp,q,r,s N gSa½ds1 vkSjLoa;dsvykokHkktdksadhla[;k gSa – Sol.

(A) p q r s (C) p q r s  2 Total number of divisors is

(B) (p + 1) (q + 1) (r + 1) (s + 1)  4 (D*) (p + 1) (q + 1) (r + 1) (s + 1)  2

dqyHkktdksadhla[;k (p + 1) (q + 1) ( r + 1) (s + 1) – 2 C-2_.

N is a least natural num ber having 24 divisors. Then the number of ways N can be resolved into two factors is (A*) 12 (B) 24 (C) 6 (D) None of these N,d,slhlclsNksVhizkd`rla[;kgSftlds24HkktdgS]rksNdksfdrusizdkjlsnksxq.ku[k.Mksads:iesaifjofrZr

fd;k tk ldrk gS& (A*) 12 Sol.





3

(C) 6

(D)buesalsdksbZugha

2

N = 2 .3 5 = 2 .3 .5 ( + 1) (  + 1) (  + 1) = 4.3.2 N = 360 = 2 3 . 3 2 .5

4.3.2 2 C-3.

(B) 24 

 12

How many divisors of 21600 are divisible by 10 but not by 15? (A*) 10 (B) 30 (C) 40

(D) none

21600 dsfdrusHkktd 10 lsfoHkkftrgSaysfdu15 lsugha ? (A) 10 Sol.

(B) 30 5

3

(C) 40


2

Here 21600 = 2 . 3 . 5 (2 × 5) × 2 4 × 3 3 × 5 1 Now numbers which are divisible by 10 = (4 + 1)(3 + 1)(1 + 1) = 40 (2 × 3 × 5) × (2 4 × 32 × 51) now numbers which are


divisible by both 10 and 15 = (4 +1)(2 +1)(1+1) = 30 So the numbers which are divisible by only 40 – 30 = 10 Hindi.

;gk¡ 21600 = 25. 33. 52 = (2 × 5) × 2 4 × 3 3 × 5 1

vcosHkktdtksfd 10 lsHkkT;gS = (4 + 1)(3 + 1)(1 + 1) = 40 21600 = (2 × 3 × 5) × (2 4 × 3 2 × 5 1)

vc os Hkktd tks fd 10o15 nksuksa ls HkkT; gSa = (4 +1)(2 +1)(1+1) = 30 vr% os la[;k;s tks fd dsoy 10ls HkkT; gS 40 – 30 = 10 C-4.

The sum of the divisors of 2 5 . 3 7 . 5 3 . 7 2 , is (A) 2 6 . 3 8 . 5 4 . 7 3

(B) 26 . 3 8 . 5 4 . 73 – 2 . 3 . 5 . 7

(C) 2 6 . 3 8 . 5 4 . 7 3 – 1

(D*) none of these

25 . 37 . 53 . 72 ds Hkktdksa dk ;ksxQy gSa & (A) 2 6 . 3 8 . 5 4 . 7 3

(B) 26 . 3 8 . 5 4 . 73 – 2 . 3 . 5 . 7

Sol.

(C) 2 6 . 3 8 . 5 4 . 7 3 – 1 (D) buesalsdksbZugha Since of the divisors of 2 5. 3 7 . 5 3 . 72 (2 0 + 2 1 + 2 2 + 2 3 + 2 4 + 2 5) (30 + 3 1 + ......+ 3 7) (5 0 + 5 1 + 5 2 + 5 3) ( 70 + 7 1 + 7 2 + 7 3)

Hindi.

ds Hkktdksa dk ;ksx 25. 37 . 53 . 72 (2 0 + 2 1 + 2 2 + 2 3 + 2 4 + 2 5) (30 + 3 1 + ......+ 3 7) (5 0 + 5 1 + 5 2 + 5 3) ( 70 + 7 1 + 7 2 + 7 3)

C-5.

The number of ways in which the number 27720 can be split into two factors which are co-primes, is:

la[;k27720 dksnkslgvHkkT;xq.ku[k.MksaesafdrusrjhdksalsfoHkkftrfd;ktkldrkgS\ (A) 15

(B*) 16

(C) 25

(D) 49

2 27720 2 13860

Sol.

5

6930

2

1386

3

693

3

231

7

77 11

23 32 . 5. 11 Hence number of co-prime factor

lgvHkkT;xq.ku[k.Mksadhla[;k 25–1 = 2 4 = 16

Section (D) : Problem based on circular arrangement / Multinomial theorem (PCMT/PCCA)

[k.M (D):o`Ùkh;Øep;rFkkcgqinh;izes;ijvk/kkfjrleL;k,¡ D-1.

The number of ways in which 8 different flowers can be strung to form a garland so that 4 particulars flowers are never separated, is: (A) 4 !. 4 !

(B)

8! 4!

(C*) 288

(D) none

8 fofHkUuQwyksals,dgkjfdrusrjhdksalscuk;ktkldrk gSa;fn4 fo'ks"kQwydHkhHkhvyxugksa\ (A) 4 !. 4 ! Sol.

(B)

8! 4!

(C) 288

(D) buesalsdksbZugha

First be find all 4 particulars flowers are together then the total num ber of ways is PERMUTATION & COMBINATION - 23

4!  4! = ........ 288 2 Hindi. 4 fo'ks"kQwyksadkslkFkj[krsgq, q,gkjcukusdsrjhdksdhla[;k=

D-2.

4!  4! = ........ 288 2

The number of ways in which 6 red roses and 3 white roses (all roses different) can form a garland so that all the white roses come together, together, is 6 ykyxqykcksavkSj3 lQsnxqykcksa¼lHkhxqykcvyx&vyxgSa½ls,dekykfdrusrjhdksalscukbZ tkldrhgSatcfd

lHkhlQsnxqykc,dlkFkjgsa\ Sol.

(A) 2170 (B) 2165 Total number of ways is

(C*) 2160

(D) 2155

dqyrjhdksadhla[;k 6!  3! = 720 × 3 = 2160 2! D-3.

The number of ways in which which 4 boys & 4 girls can stand in a circle so that each b oy and each girl is one after the other, is: 4 yM+dsvkSj4yM+fd;k¡,do`Ùkesafdrusrjhdksals [kM+sjgldrsgSarkfdizR;sdyM+dk vkSjyM+dh,dkUrjØeesajgsa\

Sol.

(A*) 3 !. 4 ! ( B) 4 !. 4 ! (C) 8 ! First we arrange all the boy so no. of ways of all the boy can stand is 3! now we arrange all the girl in 4 ! ways so total no. of ways is

(D) 7 !

= 4! × 3! Hindi.

igysgeyM+dksadksO;ofLFkrdjrsgSavr%lHkhyM+dksadks3! rjhdsls[kM+kfd;ktkldrkgSA vcyM+fd;ksadks 4 ! rjhdsls O;ofLFkrfd;k tk ldrk gS vr% dqyrjhds = 4! × 3!

D-4.

The number of ways in which 5 beads, chosen from 8 different beads be threaded on to a ring, is: (A*) 672 (B) 1344 (C) 336 (D) none 8 fofHkUueksfr;ksaesals5eksrhdkspqudj,dfjaxesafdrus usrjhdksalsyxk;ktkldrkgSa\

Sol.

(A*) 672 (B) 1344 (C) 336 First we select 5 beads from 8 different beads to 8C5 Now total number of arrangement is 8C

5 ×

4! 2!

(D)buesalsdksbZugha

= 672

Hindi. 8 fofHkUueksrh;ksaesals5eksrhpquusdsrjhds=8C5

vr%dqyfoU;klksadhla[;k 8C

5 ×

4! = 672 2! PERMUTATION & COMBINATION - 24

D-5.

Number of ways in which 3 persons throw a normal die to have a total score of 11, is 3 O;fDr,dlk/kkj.kiklsdksfdrus fdrusrjhdksalsQsadldrsgSa;fnmudk mudkdqyLdksj 11 jgs\&

Sol.

Hindi.

( A *) 27 ( B) 2 5 Using multinomial theorem Find the co-efficient of x 11 in the e xpansion (x + x 2 + x 3 + .......+ x 6)3 = x 3 (1 – x 6)3.(1–x) –3 is = 10 C 8 –3 . 4C2 = 45 – 18 = 27

(D) 18

cgqinh;izes;dkiz;ksxdjusij (x + x2 + x3 + .......+ x6)3 ds izlkj esa x11 dk xq.kkad = x3 (1 – x6)3.(1–x) –3 ds izlkj esa x11 dkxq.kkad =

D-6.

(C) 29

10

C 8 –3 . 4C2 = 45 – 18 = 27

If chocolates of a particular brand are all identical then the n umber of ways in which we can choose 6 chocolates out of 8 different brands available in the market, is:. (A*) 13C6

(B) 13C8

(C) 86

(D) none

;fnfdlhfo'ks"kczk.MdhlHkhpkdysVloZlegks]rkscktkjesamiyC/k8fofHkUuczk.Mksesals6pkdysVfdrusrjhdksals pquhtkldrhgSa\ (A) 13C6 (B) 13C8 (C) 86 (D)buesalsdksbZugha Sol.

Using multinomial theorem Total no. of ways of choosing 6 chocolates out of 8 different brand is = 8 + 6–1C 6 = 13 C6

Hindi.

cgqinh;izes;dkiz;ksxdjusij 8 fofHkUuczk.Mksesals6pkdysVpquusdsdqyrjhds= 8 + 6–1C6 = 13C6

D-7.

Number of positive integral solutions of x 1 . x 2 . x 3 = 30, is x1 . x2 . x3 = 30 ds/kukRediw.kk±dgyksadhla[;kgksxh\

Sol.

( A ) 25 ( B) 2 6 Total number of posi tive integral solution of

(C*) 27

(D) 28

/kukRediw.kk±dgyksadhdqyla[;k x 1 . x 2 . x 3 = 80 = 2 × 3 × 5 is 3 × 3 × 3 = 27

Section (E) : Problem based on geometry / Dearrangement / exponent of prime/ Principal of exculusion/Grouping (PCGT/PCDA/PCGP) (PCGT/PCDA/PCGP)

[k.M (E) :T;kfefr ij vk/kkfjr] vO;oLFkk ij] vHkkT; la[;k dh ?kkr] foHkktu dk fl)[email protected] fl)[email protected] E-1.

Number of ways in which 9 different toys be distributed among 4 children belonging to different age groups in such a way that distribution among the 3 elder children is even and the youngest one is to receive one toy more, is:

(A)

5 !

2

(B)

8

9! 2

(C*)

9! 3 ! 2 !

3

(D) none

9 vyx&vyxf[kykSusfofHkUuvk;qoxZds 4 cPpksaesafdrusrjhdksalsckaVstkldrsgS;fn3 cM+scPpksadkslela[;k

esaf[kykSusfeysvkSjlclsNksVscPpsdks,df[kYkkSukT;knkfeys\& (A)

5 ! 8

2

(B)

9! 2

(C)

9! 3 ! 2 !

3



Sol.

Here first three children receive 2 each and younger receives 3 toys then total 3 toys then total number of distribution is 9C . 7C . 5C . 3C 2 2 2 2

9! 9! 7! 5! 3! . . . = 2!. 7 ! 2!. 5 ! 2!. 3 ! 3!. 0 ! 3!.( 2! )3

=

Hindi. 3cM+scPpksaesaizR;sddksnksf[kyksusrFkklclsNksVscPpsdks3 f[kyksusnsrsgSa

vr%ckaVusdsdqyrjhdksdhla[;k

E-2*_.

9C

2 .

=

9! 9! 7! 5! 3! . . . = 2!. 7 ! 2!. 5 ! 2!. 3 ! 3!. 0 ! 3!.( 2! )3

50

7C

2 .

5C

2 .

3C

2

C36 is divisible by

C36fdllsfoHkkftrgS&

50

(B*) 52

(A*) 19 Sol.

50! exp of 19 in 50 ! = 14! 36!

50

 50   50   19  + 19 2  = 2

C36 is divisible by 19 but not by 19 2

Exp. of 5 in 50 ! =

 50    50   5   25  = 12

Exp. of 5 in 14 ! =

14  5 =2  

Exp. of 5 in 36 ! =

 36   36   5    25  = 8

Hindi. 50! esa 19 dk ?kkarkd

36! esa 19dk ?kkarkd =



(D) 53

 36   36   19    192  = 1    

Exp. of 19 in 36 ! =



(C) 192

50 ! 14! 36!

=

A ns . A & B

 50   50   19  + 19 2  = 2    

 36   36   19    192  = 1

C36, 19 lsHkkT;gSaijUrq 192 lsugha

50

 50   50   = 12  5   25 

50! esa 5 dk?kkarkd= 

14  =2  5 

14! esa 5 dk ?kkarkd= 

36! esa 5 dk ?kkarkd=

E-3*.

2n

Pn is equal to

 36   36   5    25  = 8    

A ns . A & B


Pn dkekugSa&

2n

(B*) 2n [1 . 3 . 5 .....(2n – 1)] (D*) n! (2nCn)

(A*) (n + 1) ( n + 2) ..... (2n) (C*) (2) . (6) . (10) .... (4n – 2)

Sol.

2n! n! (i) (n + 1) (n + 2) ----- (2n) Value of 2npn is

and n ! . 2nC n

1.2.3.4.5. 6.7.8.9.10 .11.....(2n – 2) (2n – 1). 2n 1.2.3........n =

(1.3.5.7.. ..(2n – 1)). (2.4.6.8.. .....2n) 1.2.3........n

=

(1.3.5.7....(2n – 1)) . 2n (1.2.3.4.......) 1.2.3.4

= 2 n (1.3.5.7. .....(2n – 1)) = (2. 6. 10. 14 ..... (4n – 2)) Hindi.

2np n

dk eku

2n! n!

gS vkSj n ! . 2nCn

(i) (n + 1) (n + 2) -- --- (2n)

1.2.3.4.5. 6.7.8.9.10 .11.....(2n – 2) (2n – 1). 2n 1.2.3........n =

(1.3.5.7.. ..(2n – 1)). (2.4.6.8.. .....2n) 1.2.3........n

=

(1.3.5.7....(2n – 1)). 2n (1.2.3.4.......) 1.2.3.4

= 2 n (1.3.5.7. .....(2n – 1)) = (2. 6. 10. 14 ..... (4n – 2))

E-4*.

There are 12 points in a plane of which 5 are collinear. The number of dis tinct quadrilaterals which can be formed with vertices at these points is:

,dleryesa12fcUnqgSaftuesals5lajs[kh;gSaAbufcUnqvksadks'kh"kZekudjcuk;stkldusokysfHkUu&fHkUuprqHkqZtksadhla[;k gksxh& Sol.

(A*) 2. 7P3 (B) 7P3 Total number of required quadrilateral

(C) 10 . 7C3

(D*) 420

dqyvHkh"VprqHkqZtksadhla[;k 7C

7C

4 +

3 ×

5C

1 +

7C

2 ×

5C

2

7 65 4 7 65 76 54   .5  1 2  3  4 1 2  3 1 2 1 2 = 35 + 175 + 210 = 420 = 2 . 7p3 =

E-5*.

The number of ways in which 200 different things can be divided into groups of 100 pairs, is: 200fofHkUuoLrqvksadks100;qXeksaesafdrusrjhdksalsfoHkkftrfd;ktkldrkgSa\ (A)

 101  102   103  200   2   2   2  ....  2 

200 !

(C*)

(B*) 

100

2

200 ! 100

2

(100) !

(D*) (1. 3. 5...... 199)


200

Sol.

=

=

C2 . 198C2 . 196 C2 ......2 C 2 100 !

200 ! 100

2

.100 !

=

101.102.103 .....200 2100

 100   102   103   100   . . .....   2   2   2   2 

AndvkSj

1.2.3.4.5.6.7.8.....200 2100 .100 ! =

=

(1.3.5.7.....199) (2. 4. 6. 8......200) 2100 .100 ! (1.3.5......199 ). 2100 .100 ! 2100 .100 !

= 1.3.5 .199

PART - I : SUBJECTIVE QUESTIONS

Hkkx- I : fo"k;kRed iz'u ¼SUBJECTIVEQUESTIONS ½ Section (A) : Problem based on Problems, Arrangements of given objects/Selection of given object (PCAD/PCSD)

[k.M (A):nhxbZoLrqvksadspquko]foU;klijvk/kkfjrleL;k,¡ 1.

A family consis ts of a grandfat her, m sons and daught ers and 2n grand children. They are to be seated in a row for dinner. The grand children wish to occupy the n seats at each end and the grandfather refuses to have a grand children on either side of him . In how many ways can the family be made to sit?

,difjokjesa,dnknkth,mcsVs&csfV;k¡vkSj2niksrs&iksfr;k¡gSAosfMujdsfy,,diafDresacSBrsgSaAiksrs&iksfr;k¡ izR;sdfljsdhnlhVksaijcSBukpkgrsgSavkSjnknkthughapkgrsfdmudsfdlhHkhrjQdksbZiksrk;kiksrhcSBsA fdrus rjhdksa ls ;g ifjokj fMuj ds fy, cSB ldrk gSa \ Sol.

Ans. (2n)! m! (m  1) First we select n grand children from 2n grand children is 2nC n Now arrangement of both group is n! × n! Now Rest all (m + 1) place where we occupy the grandfather and m sons but grandfather refuse the sit to either side of grand children so the out of m – 1 seat one seat can be selected Now required number of sitting in 2n C × n ! × n ! × (m –1) C . m ! n 1

12 n × n ! × n ! × (m –1) C 1 . m ! n!n! = 2n ! . m ! . (m –1) Hindi. 2niksrs&iksfr;ksaesaniksrsiksrhpquusdsrjhds=2nCn =

vcnksukslewgksadksO;ofLFkrdjusdsrjhd =n!×n! vccpsgq,(m+1)LFkkuksaijmcsVs&csfV;karFkknknkthcsBsxsaijUrqnknkthdsnksuksarjQdksbZHkhiksrs&iksrhugha cSBkuk gS vr% cphgqbZ(m – 1) esa ls ,d lhV pquuh gSA vr%cSBkusdsvHkh"Vrjhdksadhla[;k 2n C

n ×

n!×n!×

(m –1) C

1 .

m! PERMUTATION & COMBINATION - 28

12 n × n ! × n ! × (m –1) C 1 . m ! n!n! = 2n ! . m ! . (m –1) =

Section (B) : Problem based on Selection as well as arrangement of objects/Rank of word (PCSR/PCSI)

[k.M (B):nhxbZoLrqvksadspquko],oamudsfoU;klijvk/kkfjrleL;k,¡]fn;sx;s'kCndhojh;rkij vk/kkfjrleL;k,¡ 2.

How many five digits numbers divisible by 3 can be form ed using the digits 0, 1, 2, 3, 4, 7 and 8 if, each digit is to be used atmost one.

vadks 0,1,2,3,4,7 vkSj 8 dkmi;ksxdjrsgq, 3 lsfoHkkftrgksusokyh 5 vadksdhfdruhla[;k,¡cukbZtkldrh gSa tcfd izR;sd vad dk mi;ksx vf/kdre 1 ckj gks\ Sol.

Ans. 744 Number divisible by 3 if sum of digits divisible case-I If 1 + 2 + 3 + 4 + 8 = 18 case-II If 1 + 2 + 3 + 7 + 8 = 21 case-III If 2 + 3 + 4 + 7 + 8 = 24 case-IV If 1 + 2 + 0 + 4 + 8 = 15 case-V If 1 + 2 + 0 + 7 + 8 = 18 case-VI If 2 + 0 + 4 + 7 + 8 = 21 case-VII If 0 + 1 + 3 + 4 + 7 = 15

Number of ways = 120 Number of ways = 120 Number of ways = 120 Number of ways = 96 Number of ways = 96 Number of ways = 96 Number of ways = 96 ___________________ total number 744 Hindi. dksbZHkhla[;k3lsfoHkkftrgksxh;fnvadksadk ;ksx3lsHkkT;gksA

3.

If 1 + 2 + 3 + 4 + 8 = 18 fLFkfr-I fLFkfr-II If 1 + 2 + 3 + 7 + 8 = 21 fLFkfr-III If 2 + 3 + 4 + 7 + 8 = 24 fLFkfr-IV If 1 + 2 + 0 + 4 + 8 = 15 fLFkfr-V If 1 + 2 + 0 + 7 + 8 = 18 fLFkfr-VI If 2 + 0 + 4 + 7 + 8 = 21 fLFkfr-VII If 0 + 1 + 3 + 4 + 7 = 15

rjhdksa dh la[;k = 120 rjhdksa dh la[;k=120 rjhdksa dh la[;k=120 rjhdksadhla[;k= 96 rjhdksadhla[;k= 96 rjhdksadhla[;k= 96 rjhdksadhla[;k= 96

dqy la[;k

___________________ 744

In how many other ways can the letters of the word MULTIPLE be arranged ; (i) without changing the order of the vowels ? (ii) keeping the position of each vo wel fixed? (iii) without changing the relative order/positio n of vowels & consonants?

'kCndsv{kjksadksfdrusvU;rjhdksalsO;ofLFkrfd;ktkldrkgSaAtcfd (i) LojksadkØevifjofrZrjgsA (ii) izR;sd Loj dk LFkku fuf'pr jgsA (iii) LojksavkSjO;atuksadklkis{kØe@LFkkuvifjofrZrjgsA MULTIPLE

Sol.

Ans. (i) 3359 (ii) 59 (iii) 359 (i)Without changing the order of the vowels of MULTIPLE So we choose the first three place in 8C3 ways and the rest are arranged is

8! 5! 8!  = 3! 5! 2! 3! 2! = 3360


Hence required no. is 3360 – 1 = 3359 (ii) Keeping the position of each vowel fixed M_LT_PL_

5! = 60 2

Number of ways =

other ways = 60–1 = 59 (iii) without changing the relative order/po sition of vowels & consonants so number of ways is =

5!  3! = 60 × 6 = 360 2!

Hence required number is = 360–1 = 359 Hindi. (i) MULTIPLE esa Lojks dk Øe vifjofrZr jgs

vr%igysrhuLojkasdks8C3 rjhdslsO;ofLFkrdjsxsa rFkkckdhv{kjksadksO;ofLFkrdjsxsa vr% dqy rjhds =

8! 5! 8!  = = 3360 3! 5! 2! 3! 2!

vr% vU; rjhdks dh la[;k = 3360 – 1 = 3359 (ii) izR;sd Loj dk LFkku M_LT_PL_ fuf'pr j[kus ij dqy rjhds=

5! = 60 2

nwljs rjhds = 60–1 = 59 (iii) Lojksa o O;atuks dk lkis{k Øe@LFkku vifjofrZr j[kus ij rjhdksa dh la[;k

=

5!  3! = 60 × 6 = 360 2!

vr% vU; rjhdksa dh la[;k = 360 – 1 = 359 4.

A bouquet fr om 11 diff eren t flowers is to b e made so that it contains not less than t hree flowers. Find number of different ways of selecting flowers to form the bouquet. 11 fofHkUuQwyksadhlgk;rkls,dxqynLrkcuk;ktkrkgStcfdblesarhulsdeQwyughagksAxqynLrksadkscukus

dsfy,Qwyksadkspquusds rjhdksadhla[;kKkrdjksA Sol.

Ans. 1981 11 C 3 + 11C4 + ....... + 11C11 = 211 – 11C 0 – 11C1 – 11C2 = 1981

5.

If  = x1 x2 x3 and  = y1 y2 y3 be two three digit numbers, then find number of pairs of be formed so that  can be subtracted from  without borrowing.

 and  that can

;fn  = x1 x2 x3 rFkk  = y1 y2 y3 rhu vadksa dh nks la[;k,¡ gks] rks  o  ds dqy ;qXeksa dh la[;k Kkr dhft,ftlesa dks esalsfcukgkflyfy;s?kVk;ktkldsA Sol.

Ans.  – 

45 . (55) 2

= y1 y2 y3 – x 1 x 2 x 3 __________ Number of pairs

;qXeksa dh la[;k = (10 + 9 + 8 + + 1)2 . (9 + 8 + + 1)

= (55)2 .45



[k.M(C):nhxbZoLrqvksadsle:irFkkfHkUu&fHkUugksuslslEcfU/krleL;k 6.

Show that the number of com binations of n letters together out of 3n letters of which n are a and n are b and the rest unlike is, (n + 2). 2 n  1.

iznf'kZrdhft,fd3ni=kksa¼ftuesalsni=k,aizdkjdsgSani=k,b izdkjdsgSarFkk'ks"ki=kfHkUugSaA½esalsn i=kksads (n+2).2n1 laxzgcuk;stkldrsgSaA Sol.

a a a .......a b b b .......b all diff .          n times

n times

 n

selection of n objects = coefficient of x n in (1 + x + .....+ x n)2 (1 + x) n on solving = (n + 2). 2 n–1 Hindi.

a a a .......a b b b .......b all diff .          n times

n times

 n

n oLrqvksa dks pquus ds rjhdksa dh la[;k = (1 + x + .....+ xn)2 (1 + x)n

esa xn dk xq.kkad

gy djus ij = (n + 2). 2n–1 Section (D) : Problem based on circular arrangement / Multinomial theorem (PCMT/PCCA)

[k.M (D):o`Ùkh;Øep;rFkkcgqifn;izes;ijvk/kkfjrleL;k,¡ 7.

The integers f rom 1 to 1000 are written in order around a circle. Starting at 1, every fifteenth numbe r is marked (that is 1, 16, 31, .... etc.). This process in continued untill a number is reached which has already been marked, then find number of unmarked numbers. 1 ls1000 rd ds iw.kk±dksa dks o`Ùk ij fy[kk tkrk gSA 1 ls izkjEHk djrs gq, izR;sd 15 oha la[;k fpfUgr~ (vFkkZr~ 1, 16, 31, .... bR;kfn) dh tkrh gSA ;g izfØ;k rc rd tkjh j[kh tkrh gS tc rd fd igys ls fpfUgr~ la[;k

izkIrughagkstkrh]rksvfpfUgrla[;kvksadhla[;k Kkrdhft, Sol.

Ans. 800 In one round, marked numbers are 1, 16, 31, ..., 991 In second round marked numbers are 6, 21, 36, ..., 996 In third round marked numbers are 11, 26, 41, ..., 986 the next number will be 1 which has already been marked

  Hindi.

8.

  

67 numbers 67 numbers 66 numbers

  

67la[;k,¡

total marked numbers = 67 + 67 + 66 = 200 unmarked numbers = 1000 – 200 = 800

,d pDdj esa fpfUgr la[;k,¡  1, 16, 31, ..., 991 nwljs pDdj esa fpfUgr la[;k,¡  6, 21, 36, ..., 996 rhljs pDdj esa fpfUgr la[;k,¡  11, 26, 41, ..., 986 vxyh la[;k1 gksxh tksfd igys ls gh fpfUg~r gSA  dqy fpfUgr la[;k,¡ = 67 + 67 + 66 = 200 dqy vfpfUgr la[;k,¡ = 1000 – 200 = 800 

67la[;k,¡ 66la[;k,¡

Find number of positive integral solutions of xyz = 21600 xyz = 21600 ds /kukRed iw.kk±d gyksa dh la[;k Kkr dhft,A Ans.

Sol.

1260

xyz = 21600 = 2 5 . 3 3 . 5 2 Here if

x+y+z=5 x+y+z=3 x+y+z=2

  

7

C2 = 21

5

C2 = 10

4

C2 = 6 PERMUTATION & COMBINATION - 31

Hence required no. = 21 × 10 × 6 = 1260 Hindi. xyz = 21600 = 2 5 . 3 3 . 5 2

;gk¡;fn

x+y+z=5 x+y+z=3 x+y+z=2

  

7

C2 = 21

5

C2 = 10

4

C2 = 6

vr% vHkh"V la[;k = 21 × 10 × 6 = 1260 Section (E) : Problem based on geometry / Dearrangement / exponent of prime/ Principal of exculusion/Grouping (PCGT/PCDA/PCGP)

[k.M (E) :T;kfefr ij vk/kkfjr] vO;oLFkk ij] vHkkT; la[;k dh ?kkr] foHkktu dk fl)[email protected] 9.

The sides AB, BC & CA of a triangle ABC have 3, 4 & 5 interior points respectively on them. Find the number of triangles that can be constructed using these interior points as vertices.

f=kHkqt ABCdh Hkqtkvksa AB, BCvkSj CAij Øe'k% 3, 4 vkSj 5 vH;kUrj fcUnq (interiorpoints) fLFkr gSaA bu vH;kUrjfcUnqvksadks'kh"kZekudjfdrusf=kHkqtcuk,tkldrsgSa\ Sol.

Ans. 205 Total number of triangle = Two points taken from AB and one point either BC or CA + similarly BC + similarly  A + one po int e ach s ides . = 3C 2 [4C1 + 5C 1] + 4C2 [5C1 + 3C 1] + 5C2 [3C 1 + 4C1] + 3C 14C15C1 = 205 Total – (collinear poi nts used) = 12C 3 – (3C3 + 4C3 + 5C3) = 220 – 15 = 205

Alternate Total – Collinear points used = 12C3 – ( 3C3 + 4C3 + 5C 3) = 220 – 15 = 205 Hindi. f=kHkqtksadhdqyla[;k = nksfcUnq AB ijlsrFkk,dfcUnq BC ;kCAlsysxsa + blhizdkj BC + blhizdkjCA + ,dfcUnqizR;sdHkqtkijls = 3C 2 [4C1 + 5C 1] + 4C2 [5C1 + 3C 1] + 5C2 [3C 1 + 4C1] + 3C 14C15C1 = 205 dqy –(lajs[kh;fcUnqysdjcuk;sx;s f=kHkqt) = 12C 3 – (3C3 + 4C3 + 5C3) = 220 – 15 = 205

oSdYifd dqy – lajs[kh; ysdj cuk;s x;s f=kHkqt = 12C3 – (3C3 + 4C3 + 5C3) = 220 – 15 = 205 10.

How many positive integers of n digits exist such that each digit is 1, 2, or 3? How many of these contain all three of the digits 1, 2 and 3 atleast once ? n vadksdsfdrus/kukRediw.kk±dlEHkogSaftudkizR;sdvad1,2;k3gks\buesals,slhfdruh la[;k,¡gksxhftuesa

lHkh rhuksa vad 1, 2, 3 de ls de ,d ckj vk;s\ Ans. Sol.

3n , 3 n – 3.2 n + 3

Total n-digit numbers using 1, 2 or 3 = 3n total n-digit numbers using any two digits out of 1, 2 or 3 = 3C 2 × 2 n – 6 = 3 × 2 n – 6 PERMUTATION & COMBINATION - 32

total n-digit numbers using only one digit of 1, 2 or 3 = 3  the numbers containing all three of the digits 1, 2 and 3 at least once = 3 n – (3 × 2 n – 6) – 3 = 3n – 3 . 2 n + 3 Hindi. 1, 2 ;k3 dk iz;ksx djds dqy n vad okyh la[;k,¡= 3n 1, 2 ;k3 esa ls fdUgh nks vadksa dk iz;ksx djds dqy n vad okyh la[;k,¡= 3C2 × 2n – 6 = 3 × 2n – 6 1, 2 ;k3 esa ls dsoy ,d vad dk iz;ksx djds cuh la[;k,¡ = 3



1, 2 vkSj3 esa ls izR;sd vad dk de ls de ,d ckj iz;ksx djds cuh la[;k,¡ = 3n – (3 × 2n – 6) – 3 = 3 n – 3 . 2 n + 3

11.

Find the number of ways in which 8 non-identical apples can be distributed among 3 boys such that every boy should get atleast 1 apple & atmost 4 apples. 8 fofHkUulsc 3yM+dks esa fdrus izdkj ls ckaVs tk ldrs gSa tcfd izR;sd yM+ds dks de ls de ,d vkSj vf/kd ls

vf/kd4 lscfeyrsgSa\ Sol.

Ans. 4620 8 - non identical number of ways=

B1 B2 1 3 2 3 2 2 Here required number of ways = 3! { 8C1 . 7C3 .4C4 + 8C2 . 6C3 . =6

B3 4 3 4 3C

3

+ 8C 2 . 6C 2 . 4C 4}

 8! 8! 6! 8! 6!   3 !. 4!  2!. 6!  3 !. 3!  2 !. 6!  2!. 4!   

= 6[280 + 280 + 210] = 6 × 770 = 4620 Hindi. 8fofHkUulsc

dqyrjhdksadhla[;k=

B1 1 2 2

B2 3 3 2

B3 4 3 4

;gk¡vHkh"Vrjhdksadhla[;k = 3! { 8C1 . 7C3 .4C4 + 8C2 . 6C3 . 3C3 + 8C 2 . 6C2 . 4C4} =6

 8! 8! 6! 8! 6!   3 !. 4!  2!. 6!  3 !. 3!  2 !. 6!  2!. 4!   

= 6[280 + 280 + 210] = 6 × 770 = 4620 12.

There are ' n ' straight line in a plane, no two of which are parallel and no three p ass through the same point. Their points of intersection are joined. Show that the number of fresh lines thus introduced is,

1 8 n (n  1) (n  2) (n  3).

,dleryesa'n'ljyjs[kk,¡gSftuesalsdksbZHkh nkslekUrjughagSvkSjdksbZHkh rhujs[kk,¡lekufcUnqls ughaxqtjrhgSaAmudsizfrPNsnfcUnqvksadksfeyk;ktkrkgSaAiznf'kZrdhft,fdbl izdkjcuusokyhu;hljy js[kkvksa dh la[;k 81 n (n  1) (n  2) (n  3)gSaA Sol.

If 'n' straight line intersect each other then total


number of intersection point is nC2 =

n(n  1) 2

Now, from these nC2 points we can make

n( n1) 2 C

2

lines. (total old + new lines) and number of old lines are So fresh lines are

= Hindi.

n(n 1) 2 C

2

n 1

C2 × n

– n1C2 × n

1 n (n  1) (n  2) (n  3) 8

;fn'n' ljyjs[kk,¡,dnwljsdks izfrPNsndjrhgS]rksdqy izfrPNsnfcUnqvksadhla[;k nC2 = vcnC2 fcUnqvksalsge

n(n 1) 2 C

2

n(n  1) 2

js[kk,¡cukldrsgS

(dqyiqjkuh+u;hjs[kk;s)

RkFkkiqjkuhjs[kkvksadhla[;k n1C2 ×ngksxhA vr% u;h cuh js[kkvksa dh la[;k

n(n 1) 2 C

2

– n1C2 × n =

1 8 n (n  1) (n  2) (n  3)

PART - II : OBJECTIVE QUESTIONS

Hkkx- II : oLrqfu"B iz'u ¼OBJECTIVEQUESTIONS ½ Single choice type

,dy fodYih izdkj 1.

A train is going from Lon don to Cambridge stops at 12 intermediate station s. 75 persons enter the train during the journey with 75 different tickets of the same class. Number of different sets of tickets they may be holding is: (A*) 78C3 (B) 91C75 (C) 84C75 (D) none

yUnulsdSfEczttkusokyh,dVsªu12 ek/;fedLVs'kuksaij:drhgSaA;k=kkdsnkSjkulekuJs.khds 75 fHkUufVdVysdj75 ;k=khVsªuesaizos'kdjrsgSa]rksbufVdVksadsfdrusvyx&vyxlewgcuk,tkldrsgSa: (A) 78C3 (B) 91C75 (C) 84C75 (D)buesalsdksbZugha Sol.

Total no. of different tickets is 12 + 11 + 10 + ......+ 1 = 78 Hence required no. = 78 C75 = 78 C3

Hindi.

dqyfHkUufVdVksadhla[;k 12 + 11 + 10 + ......+ 1 = 78

vr% vHkh"V la[;k = 78C75 = 78C3 2.

In a unique hockey series betwee n India & Pakistan, they decide to play on till a team wins 5 matches. The number of ways in which the series can be won by India, if no match ends in a draw is: (A*) 126 (B) 252 (C) 225 (D) none

HkkjrvkSjikfdLrku,dgkWdhJ`a[kykrcrd[ksyrsgSatcrd fd,dVhe5 eSpughathrtkrhgSaA;fn,dHkheSpMªkugha gks]rksHkkjr;gJ`a[kykfdrusrjhdksalsthrldrkgSa\ (A) 126 (B) 252 (C) 225 (D)buesalsdksbZugha Sol.

Case -  when all 5 match win the india then total no. of ways = 5C5 PERMUTATION & COMBINATION - 34

Case -  when 6 th match win the india then total no. of ways = 5C4 Case -  when 7 th match win the india then total no. of ways = 6C4 Case - V when 8th match win the india then total no. of ways = 7C4 Case - V when 9 th match win the india then total no. of ways = 8C4 Hence required no. of ways = 1 + 5 + 6C4 + 7C 4 + 8C4 = 1 + 5 + 15 + 35 + 70 = 126 Hindi.

fLFkfr-  tclHkh5 eSpHkkjrthrtkrkgSrksdqyrjhds= 5C5 fLFkfr-  tc6 okaeSpHkkjrthrrkgS]rksdqyrjhds= 5C4 fLFkfr-  tc7 okaeSpHkkjrthrrkgS]rksdqyrjhds= 6C4 fLFkfr- V tc8 okaeSpHkkjrthrrkgS]rksdqyrjhds=7C4 fLFkfr- V tc9 okaeSpHkkjrthrrkgS]rksdqyrjhds=8C4 vr%vHkh"Vrjhds = 1 + 5 + 6C4 + 7C 4 + 8C4 = 1 + 5 + 15 + 35 + 70 = 126

3.

12 guests at a dinner party are to be seated along a circular table. Supposing that the master and mistress of the house have fixed seats opposite to one another and that there are two specified guests who must always be placed next to one another. The number of ways in which the company can be placed, is: (A*) 20. 10 !

(B) 22. 10 !

(C) 44. 10 !

(D) none

,djkf=kHkkstesa12 esgekuksadks,dxksyestijfcBk;ktkrkgSAekukfdekfydvkSjekyfduvkeus&lkeusfuf'prlhVksaij cSBrsgSvkSjnksfo'ks"kesgekulnSolkFk&lkFkcSBrsgks]rkslEiw.kZlewgdksfdrusrjhdksalsfcBk;ktkldrkgSa\ (A) 20. 10 ! (B) 22. 10 ! (C) 44. 10 ! (D)buesalsdksbZugha Sol.

First we seat first two specified person in 2×10 = 20 ways and remaining 10 person can be arranged in 10! ways.

So total no. of ways is = 2×10×10! = 20. 10 ! Hindi.

igysgenksfo'ks"kesgekuksadkscSBkrsgSatksfd2×10=20rjhdslscSBldrsgSarFkkckdhcps10O;fDr10!rjhdksalsO;ofLFkr gksldrsgSaA

vr% dqy rjhds = 2×10×10! = 20. 10 ! 4.

Six persons A, B, C, D, E and F are to be seated at a circular table. The number of ways this can be done if A must have either B or C on his right and B must ha ve either C or D on his right, is:

N%O;fDr;ksa A, B, C, D, E vkSjF dks,dxksyestdspkjksavksjfdrusrjhdksalscSBk;ktkldrkgSa;fn Adsnka;hvksjlnSo B ;kCcSBrsgSavkSjB dsnka;hvksjlnSoC ;kD cSBrsgSaA Sol.

(A) 36 (B) 12 Case- If B is right on A

(C) 24

(D*) 18


Subcase -  C is right on B then no. of ways = (4 –1)! = 6 Subcase-  If D is right on B then no. of ways = (4 –1)! = 6 Case- If C is right on A

 D must be right on B = (4 – 1)! = 3! = 6 Hence total no. of ways is 6 + 6 + 6 = 18 Hindi.

fLFkfr- ;fnB,Adsnka;hvksjgS (a) C,Bdsnka;hvksjgks]rks dqy rjhds = (4 –1)! = 6 ;fnD,Bdsnka;hvksjgks]rks dqy rjhds = (4 –1)! = 6 fLFkfr- ;fnC,Adsnka;hvksjgks  D, B dsnka;hvksjgksukpkfg, (b)

= (4 – 1)! = 3! = 6

vr% dqy rjhds 6 + 6 + 6 = 18 5.

Out of 16 players of a cricket team , 4 are bowlers and 2 are wicket k eepers. A team of 11 players is to be chosen so as to contain at least 3 bowlers and at least 1 wicketkeeper. The number of ways in which the team be selected, is

,dfØdsVVheds16 f[kykfM+;ksaesa4xsancktvkSj2 fodsVdhijgSA11 f[kykfM+;ksadh,dVhefdrusrjhdkslscuk;htkldrh gSa]ftlesadelsde3 xsancktvkSjdelsde,d fodsV&dhijgks\ (A) 2400 (B*) 2472 (C) 2500 Sol. Number of bowlers = 4 Number of wicketkeeper = 2 Let total number required selection 4C .2C .10 C + 4C .2C .10 C +4C . 2C .10 C + 4C . 2C .10 C 3 1 7 4 1 6 3 2 6 4 2 5 960 + 420 + 840 + 252 = 2472 Hindi. xsanksadhl[;k=4

(D) 960

fodsVdhijksadhla[;k = 2 vHkh"Vpqukoksadhla[;k 4C

3 .

2C

1 .

10 C

7 +

4C

2 10 4 2 10 4 2 10 4 . C 1 . C 6 + C 3 . C2 . C6 + C4 . C2 . C 5

960 + 420 + 840 + 252 = 2472 6.

The number of ways in which 15 identical apples & 10 identical oranges can be distributed am ong three persons, each receiving none, one or more is: (A) 5670

(B) 7200

(C*) 8976

(D) none of these PERMUTATION & COMBINATION - 36

15 le:ilscvkSj10le:ilarjksdksrhuO;fDr;ksaesafdrusrjhdksalsckaVktkldrkgS;fnizR;sdO;fDrdks'kwU;],d;k

vf/kdfeysA (A) 5670 Sol.

(B) 7200

17  16 12  11  = 8976 2 2

cgqinh;izes;dkiz;ksxdjusij dqy rjhds = 15 + 3 – 1C15 × 10 + 3 – 1C10 = =

7.

(D)buesalsdksbZugha

Using multinomial theorem Total no. of ways = 15 + 3 – 1C 15 × 10 + 3 – 1C10 = 17 C15 × 12 C 10 =

Hindi.

(C) 8976

17

C15 × 12 C 10

17  16 12  11  = 8976 2 2

The number of perm utations which can be formed out of the letters of the word "SERIES" taking three letters together, is: (A) 120 (B) 60 (C*) 42 (D) none SERIES 'kCndsv{kjksaesals3 v{kjksadks,dlkFkysdjcuk,tkldusokysØep;ksadhla[;kgS:

Sol.

(A) 120 (B) 60 SERIES S - 2, E - 2 , R,  case-I when all letter distinct is 4C × 13 = 4 × 6 = 24 3 case-II when 2 letters are same the 2C

3 1 . C 1 ×

(C) 42

(D)buesalsdksbZugha

4! = 2 . 3 . 3 = 18 2!

total number is 24 + 18 = 42 Hindi. SERIES S - 2, E - 2 , R, 

fLFkfr-I tclHkhv{kjfHkUugSaA 4C

3 ×

13 = 4 × 6 = 24

fLFkfr-II tc2v{kjlekugSaA 2C

3 1 . C 1 ×

4! 2!

= 2 . 3 . 3 = 18

dqy la[;k,¡ 24 + 18 = 42 8.

Seven different coins are to be divided amongst thr ee persons. If no two of th e persons receive the same number of coins but each receives atleast one coin & none is left over, then the number of ways in which the division may be made is: (A) 420 (B*) 630

(C) 710

(D) none

7fofHkUuflDdsrhuO;fDr;ksaesackaVstkrsgSaA;fnfdUghHkhnksO;fDr;ksadkslekula[;kesaflDdsughafeyrsgSaysfduizR;sd

dksdelsde,dflDdkfeyrkgSvkSjdksbZflDdk'ks"kughajgrkgS]rksbuflDdksadksfdrusrjhdksalsckaVktkldrkgSa: (A) 420 (B) 630 (C) 710 (D)buesalsdksbZugha Sol.

Coin dividing in any are possible i.e. 1, 2, 4 so the number of ways is 7C . 6C . 4C . 3! = 7 × 15 × 6 = 630 1 2 4


Hindi.

flDdsfuEurjhdkslsckaVstkldrsgSavFkkZr~1,2,4 vr%dqyrjhdksadhla[;k 7C

9.

1 .

6C

2 .

4C

4 .

3! = 7 × 15 × 6 = 630

The streets of a city are arranged like the lines of a chess board. There are m streets running North to South & 'n' streets running East to West . The number of ways in which a man can travel from NW to SE corner going the shortest possible distance is:

,d'kgjdhxfy;ksadks'krjatdhfclkrdhjs[kkvksadhrjgO;ofLFkrfd;kx;kgSAmÙkjlsnf{k.kdhvksjtkusokyhm xfy;k¡ gSvkSjiwoZlsif'pedhvksjtkusokyhn xfy;k¡gSA,dO;fDrmÙkj&if'pedksuslsnf{k.k&iwoZdksusrdfdrusrjhdksalstk ldrkgSatcfdogU;wurelEHkkfornwjhlstk;sA (A) Sol.

m2  n 2

(C)

( m  n) ! m! . n!

(D*)

(m  n  2) ! (m  1) ! . ( n  1) !

(m  n  2) ! (m  1) ! . ( n  1) !

gesa(n –1) xfy;kaiwoZdhvksjrFkk(m–1) xfy;kanf{k.kdhvksjr;djuhgS vr% dqy pyh x;h xfy;ka = (m + n – 2) vr%dqyrjhds =

10.

(m  1) 2 . ( n  1) 2

Here we should go (n –1) steps to east and (m –1) steps to south so total steps which we have to g o are (m + n – 2) ways. Hence total no. of ways = m + n – 2Cm–1 . n –1 Cn–1 =

Hindi.

(B)

m+n–2

C m–1 . n –1 C n–1 =

(m  n  2) ! (m  1) ! . ( n  1) !

The number of ways in which a mixed double tennis game can be arranged from am ongst 9 married couple if no husband & wife plays in the same gam e is: 9 fookfgr;qxyksadksfefJr;qxyVsfulfdrusizdkjlsf[kyk;ktkldrkgS;fndksbZHkhifrvkSjiRuh,d gh[ksy(game)

esaugha[ksyldrsgSa? Sol.

(A) 756 (B) 3024 (C*) 1512 (D) 6048 There are 9 married couple so first we select 2 man out of 9 and then we select 2 women out of rest 7 then, we arranged them, so required no. is 9C 2 × 7C2 × 2! = 36 × 21 × 2 = 1512

Hindi. 9 fookfgr;qxygSvr%igysge9 esals2 iq:"kksadkp;udjsaxsarFkkckdh7 efgykvksaesals2dkp;udjsaxsa

rFkkbUgsaO;ofLFkrdjsxsa]vr%vHkh"Vla[;k 9

11.

C 2 × 7C 2 × 2! = 36 × 21 × 2 = 1512

The number of ways in which 5 X's can be placed in the squares of the figure so that no row remains empty is:

fn;sx;sfp=kdsoxks±esaik¡pX fdrusizdkjlsj[kstkldrsgSarkfddksbZHkhiafDr[kkyhujgs\

(A) 97 (B*) 44 (C) 100 8 Sol. Total numb er is C5 = 56 not required is 6C 5 + 6C 5 = 12 Hence required no of arrangement = 56 – 12 = 44 Hindi. dqy la[;k,¡ 8C5 = 56

(D) 126


tksvko';dugha]murjhdksadhla[;k 6C5 + 6C5 = 12 vr%vHkh"VfoU;klksadhla[;k = 56 – 12 = 44 12.

In a conference 10 speakers are present. If S 1 wants to speak before S 2 & S2 wants to speak after S3, then the number of ways all the 10 speakers can give their speeches with the above restriction if the remaining seven speakers have no objection to speak at any number is :

,dlEesyuesa10oDrkmifLFkrgSaA;fnoDrkS1 , S2 lsigyscksyukpkgrkgSvkSjS2 , S3dsckncksyukpkgrkgSvkSj;fn 'ks"klkroDrkfdlhHkhLFkkuijcksyldrsgSa]rkslHkh10oDrkviusHkk"k.kfdrusizdkjlsnsldrsgSa\ (A) 10C3 Sol.

(D*)

10 ! 3

10 C

3 .

7.2 ! =

10! 3

loZizFkege10oDrkvksaesals3oDrkvksadkp;udjsxsarFkkmUgsafdlhHkhrjhdslsj[knsxsavccpsgq,soDrkvksadsfy,dksbZ Hkhizfrca/kughagksxkA vr%dqyrjhdksadhla[;k =

13.

(C) 10P3

First we select 3 speaker out of 10 speaker and put in any way and rest are no restriction i.e. total number of ways =

Hindi.

(B) 10P8

10 C

3 . 7.2 ! =

10! 3

If all the letters of the word "QUEUE" are arranged in all possible manner as they are in a dictionary, then the rank of the word QUEUE is:

;fn'kCn"QUEUE" dsv{kjksalsfufeZrlHkh'kCnksadks'kCndks"kdsØekuqlkjO;ofLFkrfd;ktk;s]rks'kCnQUEUE dkØe gS: Sol.

(A) 15th (B) 16 th Word QUEUE E  2, Q , U - 2

(C*) 17th

(D) 18th

= 18

4! 2! = 3

3! 2! = 1 17th rank

= 1 14.

Two variants of a test paper are distributed among 12 students. Number of ways of seating of the students in two rows so that the students sitting side by side do not have identical papers & those sitting in the same column have the same paper is:

,diz'ui=kdsnksdksM12fo|kfFkZ;ksaesack¡VstkrsgSa]rksbufo|kfFkZ;ksadksnksiafDr;ksaesafdrusizdkjlsfcBk;ktkldrkgSfd izR;sdiafDresaikl&iklcSBsfo|kfFkZ;ksadkslekuiz'u&i=kughafeysrFkkizR;sdLrEHkdslHkhfo|kfFkZ;ksadkslekuiz'ui=kfeysA (A)

12! 6! 6 !

(B)

(12)! 25 . 6!

(C) (6 !) 2. 2

(D*) 12 ! × 2


Sol.

= total number of required possible is

12 C

=

15.

6

×6!×6!×2!

12! × 6 ! × 6 ! × 2 ! = 2 × 12 ! 6 ! 6 !

Sum of all the numbers that can be f ormed using all the digits 2, 3, 3, 4, 4, 4, is:

vadks2,3,3,4,4,4 dks,dlkFkiz;qDrdjrsgq,fufeZrla[;kvksadk;ksxgksxk – (A*) 22222200 Sol.

(B) 11111100 =

5! = 10 2! 3 !

=

5! = 20 3!

=

5! = 30 2! 2 !

(C) 55555500

(D) 20333280

Hence sum of unit places is 2 × 10 + 3 × 20 + 4 × 30 = 200 Hence required sum is = 200 × (105 + 10 4 + 1 03 + 102 + 101 + 10 0) = 200 × (111111) = 2222220 0 Hindi.

=

5! = 10 2! 3 !

=

5! = 20 3!

=

5! = 30 2! 2 !

vr%bdkbZdsLFkkuijfLFkrvadksdk;ksx 2 × 10 + 3 × 20 + 4 × 30 = 200

vr%vHkh"V;ksx = 200 × (105 + 10 4 + 1 03 + 102 + 101 + 10 0) = 200 × (111111) = 2222220 0 16.

Six married couple are sitting in a room. Number of ways in which 4 people can be selected so that there is exactly one married couple among the four is:

,ddejsesaN%fookfgr;qxycSBsgq,gSaAbuesalspkjO;fDrfdrusrjhdslspqustkldrsgS\rkfdbupkjksaesaBhd,dfookfgr ;qxyvk,A Sol.

Hindi.

(A*) 240 (B) 255 (C) 360 First we select one married couple out of 6 married couple i.e. 6C ways 1 total number of required case 6C × 5C × 4C × 2 = 6 × 5 × 4 × 2 = 240 1 1 1

loZizFkege6 fookfgr;qxyesals,dfookfgr,d ;qxydkp;udjsxsa6C1 vr%dqyvHkh"Vrjhdksadhla[;k 6C

17.

(D) 480

1

× 5C1 × 4C1 × 2 = 6 × 5 × 4 × 2 = 240

The number of ways selecting 8 books from a library which has 10 books each of Mathematics, Physics,


Chemistry and English, if books of the same subject are alike, is:

,diqLrdky;esaxf.kr]HkkSfrdh]jlk;u'kkL=kvkSjvaxzsthizR;sddh10-10 iqLrdsagSaA;fn,dfo"k;dhlHkhiqLrds,dleku gks]rksiqLrdky;esals8 iqLrdsap;udjusdsdqyrjhdsagSa& Sol.

Hindi.

(A) 13C4 (B) 13C3 Using multinomial theorem total number of req uired selection is 8+3 C = 11C = 11C 8 8 3

(D*) 11C3

cgqifn;izes;dsmi;ksxls dqyvHkh"Vp;udsrjhdksadhla[;k 8+3 C

18.

(C) 11C4

8 =

11C

8 =

11C

3

The number of integers which lie between 1 and 106 and which have the sum of the digits equal to 12 is: 1 vkSj106 dschpfdrusiw.kk±dfLFkrgSaftudsvadksdk;ksx12 gSa:

Sol.

Hindi.

(A) 8550 (B) 5382 (C*) 6062 Let number be x 1 x 2 x 3 x 4 x 5 x 6 But Here x 1 + x 2 + .... x 6 = 12 so coefficient of x 12 in expansion (1 + x + x 2 + .... + x 9)6 = (1 – x 10)6 . (1 – x) –6 17 C – 6C . 7C  12 1 2 = 6188 – 126 = 6062

ekuk la[;k x1 x2 x3 x4 x5 x6 gSA ijUrq ;gka x1 + x2 + .... x6 = 12 (1 + x + x2 + .... + x9)6 ds izlkj esa x12 dk xq.kkad vr% = (1 – x 10)6 . (1 – x) –6 17 C – 6C . 7C 12 1 2 = 6188 – 126 = 6062



19.

(D) 8055

In a shooting competition a man can score 0, 2 or 4 points for each shot. Then the number of different ways in which he can score 14 points in 5 shots, is: (A) 20

(B) 24

(C*) 30

(D) none

,dfu'kkusckthizfr;ksfxrkesa,dO;fDrizR;sdfu'kkusdsfy,0,2;k4vadizkIrdjldrkgSaAog5fu'kkuksaesa14vadfdrus rjhdksalsizkIrdjldrkgSa\ (A) 20 (B) 24 (C) 30 (D)buesalsdksbZugha Sol.

Find the coefficient of x 14 in the expansion of (x 0 + x 2 + x 4)5 = (1 + x 2 + x 4)5 5

 1  x 6   6 5 2 –5 =  2  = (1 – x ) (1 – x ) 1 x    = (1 – 5x 6 + 10x 12 ............) 1  5C1x 2  6C2 x 4  7 C3 x 6  ........ = =

11

C7 – 5 . 8C4 + 10.5

11 10  9  8 1 2  3  4

 5.

87 65 + 50 1 2  3  4

= 330 – 350 + 50 = 30 Hindi. (x0 + x2 + x4)5 = (1 + x2 + x4)5

ds izlkj esa x14dk xq.kkad Kkr djus ij

5

 1  x 6   6 5 2 –5 =  2  = (1 – x ) (1 – x )  1 x   = (1 – 5x6 + 10x12............) 1  5C1x 2 = =

 6C2 x 4  7 C3 x 6  ........ ds izlkj esa x14 dk xq.kkad

11

C7 – 5 . 8C4 + 10.5

11 10  9  8 1 2  3  4

 5.

87 65 1 2  3  4

+ 50 PERMUTATION & COMBINATION - 41

= 330 – 350 + 50 = 30 20.

Number of ways in which a pack of 52 playing cards be distributed equally among four players so that each may have the Ace, King, Queen and Jack of the same suit, is:

36 ! . 4 !

36 !

(A)

(B*)

4

  9!

9 !

(C)

4

36 !

(D) none

4

9 ! . 4 !

rk'kds52 iÙkksadks4O;fDr;ksaesaleku:ilsfdrusrjhdksalsckaVktkldrkgSatcfdizR;sdO;fDrdkslekuizdkj(same suit) dsbDdk]ckn'kkg]csxevkSjxqykefeyrsgks& 36 !

(A) Sol.

9 !

(C)

4

36 !

9 !

4

(D)buesalsdksbZugha

. 4!

9 .

27 C

9 .

18 C

9 9 . C9 . 4 ! =

36 ! (9! )4

×4!

vHkh"Vrjhdksadhla[;k 36 C

21.

4

36 ! . 4 !

Required number of ways 36 C

Hindi.

9 !

(B)

9 .

27 C

9 .

18 C

9 9 . C9 . 4 ! =

36 ! (9! )4

×4!

A box cont ains 6 balls which ma y be all of different colou rs or three each of two colours or two eac h of three different colours. The number of ways of selecting 3 balls from the box (if ball of same colour are identical), is: (A) 60 (B*) 31 (C) 30 (D) none

,dcDlsesa6 xsansftuesalslHkhvyx&vyxjaxdhgksldrhgSa;knksjaxdhrhu&rhuxsansgksldrhgS;krhujaxdhnks&nks xsansgksldrhgSaA¼;fnlekujaxdhxsansloZlegks½]rkscDlsesals3xsanspquusdsrjhdsgSa – (A) 60 (B) 31 (C) 30 (D)buesalsdksbZugha Sol.

Case -I If all are different then no. of ways is 6 = C 3 = 20 Case-II If three each of two colours, then combination is 2 1 0 3!  1 1 1 1! = 3! + 1! = 7ways  Case-III If two each of three colours, then combination is 3 0  2! 2 1 = 2! + 2! = 4 ways  2! Hence required no.is = 20 + 7 + 4 = 31

Hindi.

fLFkfr-I ;fnlHkhfHkUugks]rksdqyrjhds= 6C3 = 20 fLFkfr-II ;fnnksjaxdhrhu&rhuxsansgks]rkspquusdsrjhds 2

1

0

1

1

1

 

3! 1!

= 3! + 1! = 7 rjhds

fLFkfr-III ;fnrhujaxksadhnks&nksxsansagks]rkspquusdsrjhds 3

0

2

1

 2!  2!

= 2! + 2! = 4 rjhds

vr% vHkh"V la[;k = 20 + 7 + 4 = 31 22.

Number of ways in which 2 Indians, 3 Americans, 3 Italians and 4 Frenchmen can be seated on a circle, if the people of the same nationality sit together, is: (A) 2. (4 !) 2 (3 !) 2 (B *) 2. (3 !) 3. 4 ! (C) 2. (3 !) (4 !) 3 (D) none PERMUTATION & COMBINATION - 42

2 Hkkjrh;] 3 vesfjdu]3 bVkfy;uvkSj 4 ÝsUp,do`ÙkesafdrusrjhdkslscSBldrsgSa;fnlekujk"Vªh;rkds

O;fDr,dlkFkcSBukpkgrsgks\ (D) buesalsdksbZugha

(A) 2. (4 !) 2 (3 !) 2 (B) 2. (3 !) 3. 4 ! (C) 2. (3 !) (4 !) 3 Sol. Indians - 2 Ame rica ns - 3 Italians - 3 Frenchmen - 4 total number arranging in row of same nationality are together = 3 ! × 2! × 3! × 3 ! × 4 ! = 3 ! × 2! × 3! × 3 ! × 4 ! = 2 . (3!) 3 . 4! Hindi. Hkkjrh;- 2

vesfjdu- 3 bVkfy;u- 3 ÝsUp- 4 leku jk"Vªh;rkds O;fDr dks,d lkFk j[kdj cSBkus ds rjhds = 3 ! × 2! × 3! × 3 ! × 4 ! = 3 ! × 2! × 3! × 3 ! × 4 ! = 2 . (3!) 3 . 4! 23.

Number of ways in which all the letters of the word " ALASKA " can be arranged in a circle distinguishing between the clockwise and anticlockwise arrangement, , is: (A) 60 (B) 40 (C*) 20 (D) none of these " ALASKA"'kCndsv{kjksadks,do`ÙkesafdrusrjhdksalsO;ofLFkrfd;ktkldrkgStcfdnf{k.kkorZvkSjokekorZ

Øe dks vyx&vyx ekuk tkrk gSa \ (A) 60

Sol.

(B) 40

A L A S K A

Total number of arrangement is

dqy foU;klksa dh la[;k 24.

(D)buesalsdksbZugha

(C) 20

5! 120 = = 20 3! 6

5 ! 120 = = 20 3! 6

Let Pn denotes the number of ways of select ing 3 people out of ' n ' sitting in a row, if no two of them are consecutive and Q n is the corresponding figure when they are in a circle. If P n  Qn = 6, then ' n ' is equal to:

ekukfdPn ,diafDresacSBsgq,'n'O;fDr;ksaesals3 O;fDr;ksadsp;udjusdsdqyrjhdksadksiznf'kZrdjrkgStcfdmuesa lsdksbZHkhnksØekxrughagSvkSjQn mulaxrrjhdksadksiznf'kZrdjrkgStcfdos,do`ÙkesacSBsgksA;fnPn Qn = 6 gks] rks'n'cjkcjgksxk& (A) 8 Sol.

(B) 9

Here;gk¡

 

(C*) 10

(D) 12

Pn =

x 1 + x 2 + x 3 + x 4 = n – 3 x 1 + y2 + y3 + x 4  n – 5



x 1 , x 4 x 2 , x 3

 0  1 PERMUTATION & COMBINATION - 43

pn =

n – 5 + 4 – 1c

3 =

n – 2c

3

y1 , y2

 0

Q n =

 

x 1 + x 2 + x 3 = n – 3 y1 + y2 + y3 = n – 6 Now vc Pn – Q n = 6



(n – 2) (n – 3) (n – 4) n(n – 4) (n – 5) – =6 1.2.3 1.2.3 n = 10

More than one choice type

cgqfodYihizdkj 25.

The number of non-negative integral solutions of x 1 + x2 + x3 + x4  n (where n is a positive integer) is x1 + x2 + x3 + x4  n (tgk¡ n ,d /kukRed iw.kk±d gSa) ds v_.kkRed iw.kk±d gyksa dh la[;k gSa &

(A) n+3C3 (B*) n+4 C4 Sol. x 1 + x 2 + x 3 + x 4  n  x 1 + x2 + x3 + x4 + y = n (where y is known as pseudo variable) Total no. of required sol ution is = n + 5 –1C n = n+4 Cn or n+4 C4 Hindi. x 1 + x 2 + x 3 + x 4  n  x 1 + x2 + x3 + x4 + y = n

(C) n+5C5

(D*) n+4 Cn

(;gk¡ y ,d Nn~e pj gS)

dqyvHkh"Vgyksadhla[;k = 26.

n + 5 –1

C n =

n+4

Cn =

n+4

C4

A student has to answer 10 out of 13 question s in an examination. The number of ways in which he can answer if he must answer atleast 3 of the first five questions is:

,dfo|kFkhZdks,dijh{kkesa13iz'uksaesals10iz'uksadsmÙkjnsusgSaAogfdrusrjhdksals mÙkjnsldrkgS;fn mls izFke 5 iz'uksaesalsdels de 3 dsmÙkjnsukvko';dgks\ Sol.

27.

(A*) 276 (C*) 13C10 – 5C3 Total number of required poss ibilities 5C . 8C + 5C . 8C + 5C . 8C . 5C 3 7 4 6 5 5 5 = 5C 3 . 8C 7 + 5C4 . 8C6 + 8C 6 = 13C 10 – 5C 3 = 276

(B) 267 (D*) 5C3 . 8C 7 + 5C 4 . 8C 6 + 8C 5

You are given 8 balls of differe nt colour (black , white,...). Th e number of ways in which these balls can be arranged in a row so that the two balls of particular colour (say red & white) may never come together is: (A*) 8 !  2.7 ! (B*) 6. 7 ! (C*) 2. 6 !. 7C2 (D) none

fofHkUujaxks(dkyh]lQsn]yky...)dh8xsanksdks,diafDresafdrusrjhdksalsO;ofLFkrfd;ktkldrkgS;fnfo'ks"k jaxdhnksxsans¼ekukfdykyvkSjlQsn½dHkhHkh,d lkFkuvk;sa\ PERMUTATION & COMBINATION - 44

(A) 8 !  2.7 ! Sol.

(B) 6. 7 !

(D) buesalsdksbZugha

(C) 2. 6 !. 7C2

Required number of possible is

dqyvHkh"VlaHkkoukvksadhla[;k 8 ! – 2. 7 ! = 7 ! (8 – 2) = 6 . 7! 2 . 6 ! . 7C 2 28.

There are 10 seats in the first row of a theatre of which 4 are to be occupied. The number of ways of arranging 4 persons so that no two persons sit side by side is:

,dflusek?kjdhizFkeiafDresa10lhVksaijpkjO;fDr;ksadksfdrusrjhdksalscSBk;ktkldrkgS]tcfddksbZHkh nksO;fDrikl&iklucSBsa\ (A) 7C4

(B*) 4. 7P3

(C*) 7C 3. 4 !

(D*) 840

Sol. x 1 + x 2 + x 3 + x 4 + x 5 = 6 x 1 + y1 + y2 + y3 + x 5 = 3 but ijUrq x1, x5  0 x 2, x 3, x 4  1 y1, y 2, y 3  0 

vr%vHkh"Vrjhdksadhla[;k 3 + 5 – 1C

4 ! = 7C3 . 4 ! = 7p3 . 4 = 840 29.

3 .

Number of ways in which 3 numbers i n A.P. can be selected from 1, 2, 3,...... n is:

(n  2)(n  4) (A) if n is even 4

n  1 (C*)

(B)

n2

2

2

4

if n is odd

 4n  5

(D*)

n  n  2 4

if n is odd

if n is even

la[;kvksa1,2,3,......n esalslekUrjJs<+hesa 3 la[;k,¡fdrusizdkjlspquhtkldrh gSa\ (A)

(n  2)(n  4) 4

;fn n le gSa

(B)

2

 n  1 ;fnn fo"kegSa (C) 4

Sol.

(D)

n2

 4n  5 2

n  n  2 4

;fnn fo"kegSa

;fn n le gSa

Here given no. be 1,2,3,.........n Let common difference = r Total way of selec tion = (1, 1 + r, 1+2r),(2, 2 + r, 2 + 2r), ..(n – 2r, n – r, n) Total numbers are = (n – 2r) Here r min. = 1 and r max. = (n – 1)/2 Case- I When n is odd

 r max =

(n  1) 2

& total no. of selection is

(n1) / 2

=

(n  2r ) r 1


 n  1  n  1  2   n( n  1) 2  2   n  1    = – =  2 2  2  2

Case - II when n is even = r max =

n2 2

so total no. selection is

 n  2  n 2  n(n  2) (n  2r ) = –  2  2 2 r 1 2

(n2) / 2



=

= Hindi.

 n  2   n     n   = n(n  2)  2   2  4

nh x;h la[;k,¡ 1,2,3,.........n ekuk fd lkoZvUrj = r pquusdsdqyrjhds = (1, 1 + r, 1+2r),(2, 2 + r, 2 + 2r), ..(n – 2r, n – r, n)

dqy la[;k,¡ = (n – 2r) ;gk¡ r min. = 1 rFkk r max. = (n – 1)/2 fLFkfr- I tc n fo"ke gks  r max =

(n  1) 2

rFkk pquus ds dqy rjhds

(n1) / 2

=

(n  2r ) r 1

n( n  1) = – 2

 n  1  n  1     2  2 

2

2

 n  1   =   2 

fLFkfr - II tc n le gks= r max =

2

n2 2

vr% pquus ds dqy rjhds (n2) / 2

(n  2r ) = n(n2 2) –

=

 n  2  n   2  2

2

2

r 1

=

30.

 n  2   n     n   = n(n  2)  2   2  4

The number of ways in which 10 students can be divided into three teams , one containing 4 and others 3 each, is 10fo|kfFkZ;ksadks3VheksaesafdrusizdkjlsfoHkkftrfd;ktkldrkgSa;fn,dVheesa4fo|kFkhZgSavkSj'ks"knksesa

izR;sdesa3fo|kFkhZgks\ (A) Sol.

10 ! 4!3! 3!

(B*) 2100

(C*) 10C4 . 5C3

(D)

10 ! 1 . 6!3!3! 2

Total required number of te ams is


10

= = Hindi.

C 4 . 6 C3 . 3 C 3 3!

10

10! 6!  3 4!  6! 3 ! 3!  3!

 3 =

C 4 . 5C 2 = 2100

dqyvHkh"VVheksadhla[;k 10

= =

C 4 . 6 C 3 . 3C 3 3!

10

10! 6!  3 4!  6! 3 !  3!  3!

 3 =

C 4 . 5C 2 = 2100

PART - I : MATCH THE COLUMN

Hkkx- I : dkWye dks lqesfyr dhft, (MATCH THE COLUMN) 1.

Consider the word "HONOLULU". Column – I (A) Number of words that can be formed using the letters of the given word in which consonants & vowels are alternate is

Column – II (p) 26

(B)

Number of words that can be formed without changing the order of vowels is

(q)

144

(C)

Number of ways in which 4 letters can be selected from the letters of the given word is

(r)

840

(D)

Number of words in which two O's are together but U's are separated is

(s)

900

Ans.

(A)  (q),

(B)  (r),

(C)  (p),

(D)  (s)

ekuk'kCn"HONOLULU" gSA LrEHk – I (A) fn,x,'kCndsv{kjksadsmi;ksxlscuusokysmu'kCnksadhla[;kftuesa O;atu,oaLoj,dkUrjØeesavkrsgks]gS&

Sol.

LrEHk – II (p)

26

(B)

LojksadkØefcukifjofrZrfd,cuusokys'kCnksdhla[;kgS&

(q)

144

(C)

fn,x,'kCnls4 v{kjksadkmi;ksxdjdscuk,tkusokys'kCnksdhla[;kgS&

(r)

840

(D)

'kCnksadhla[;kftleasnksO's lkFk&lkFkysfduU's vyx&vyxgks]gS&

(s)

900

(A)

2!

(B)

8C

(C)

Case-I

4! 4! 2! 2! 2! = 144

4

4! = 840 2!

[P & C, E]

[P & C, E] 2-alike + 2-alike = 3C2

[P & C, E]


2-alike + 2 diff. = 3C 1 4C2 All diff. = 5C4 = 3 + 18 + 5 = 26

Case-II Case-III

(D)

5! 2!

Hindi. (A)

2!

6C

2 =

900

[P & C, E]

4! 4! 2! 2! 2! = 144

4! = 840 2!

(B)

8C

(C)

fLFkfr-I fLFkfr-II fLFkfr-III

4

[P & C, E]

[P & C, E] 2-leku + 2-leku = 3C2 2-leku+ 2 fHkUu

[P & C, E]

= 3C1 4C2

lHkhfHkUu=5C4 = 3 + 18 + 5 = 26

(D) 2.

5! 2!

6C

2 =

900

[P & C, E]

Column – I

Column – II

(A)

The total number of selections of fruits which can be made from, 3 bananas, 4 apples and 2 oranges is, it is given that fruits of one kind are identical

(p)

Greater than 50

(B)

If 7 points out of 12 are in the same straight line, then the number of triangles formed is

(q)

Greater than 100

(C)

The number of ways of selecting 10 balls from unlimited number of red, black, white and green balls is, it is given that balls of same colours ar e identical

(r)

Greater than 150

(D)

The total number of proper divisors of 38808 is

(s)

Greater than 200

Ans.

(A)  (p),

(B)  (p, q, r), (C)  (p, q, r, s),

(D)  (p)

LrEHk –

LrEHk –

(A)

3dsys]4lscvkSj2lUrjksaesalsQypquusdsdqyrjhds]

(p)

50lsT;knk

(B)

tcfd,dizdkjdslHkhQylekugSa]gksaxs& ;fndqy12fcUnqvksaesals7fcUnq,dghljyjs[kkijgS] rksbuls cuusokysf=kHkqtksadhla[;kgS&

(q)

100lsT;knk

(r)

150lsT;knk

(D)

vlhferla[;kesamifLFkryky]dkyh]lQsnvkSjgjhxsnksaesals 10xsanspquusdsrjhdksadhla[;k]tcfdlekujaxdh lHkhxsnsa loZlegaS]gksxh& la[;k 38808 ds mfpr Hkktdksa(properdivisors) dh dqy la[;k gS &

Ans.

(A)  (p),

(C)

(B)  (p, q, r), (C)  (p, q, r, s),

(s)

200lsT;knk

(D)  (p)

Sol.

[P & C, T] (A) (B)

Required number of ways = (2 + 1) (3 + 1) (4 + 1) – 1 = 59. The number of ways of selecting 3 points out of 12 points is 12C3 . Three points out of 7 collinear points can be selected in 7C3 ways. Hence, the number of triangles formed is 12C3 – 7C3 = 185.

(C)

Requied number of ways


= Coefficient of x 10 in (1 + x + x 2 + ......)4 = Coefficient of x 10 in (1 – x) –4 = 10+4–1C4–1 = 13C3 = 286 (D)

Hindi

(A)

Factorizing the given number, we have 38808 = 23 . 32 . 72 . 11 The total number of divisors of this number is same as the num ber of ways of selecting some or all of two 2’s, two 3's, two 7’s and one 11. Therefore, the total number of divisors = (3 + 1) (2 + 1) (2 + 1) (1 + 1) = 72 Hence, the required number of divisors = 72 – 2 = 70 dqy vHkh"V rjhds = (2 + 1) (3 + 1) (4 + 1) – 1 = 59.

(B)

12fcUnqvksaesals3fcUnqp;udjusdsrjhds=12C3 . 7lajs[kh;fcUnqvksaesals3fcUnq7C3rjhdslspqustkldrsgSaA

vr%cuusokysf=kHkqtksadhla[;k=12C3 –7C3 =185. (C)

vHkh"Vrjhds = (1 + x + x2 + ......)4 esa x10 dk xq.kkad = (1 – x) –4 esa x10 dk xq.kkad = 10+4–1C4–1 = 13C3 = 286

(D)

nhx;hla[;kdsxq.ku[kaMdjusij 38808 = 23 . 32 . 72 . 11

nhx;hla[;kdsHkktdksadhla[;koghagksxhtksfdnks 2’s,nks3's,nks 7’srFkk,d11esalsdqN;klHkhdks p;udjusdsrjhdsgks = (3 + 1) (2 + 1) (2 + 1) (1 + 1) = 72 vRk% vHkh"V mfpr Hkktdksa dh la[;k =72–2=70

PART - II : COMPREHENSION

Hkkx- II : vuqPNsn (COMPREHENSION) Comprehension # 1 There are 8 official and 4 non-official members, out of these 12 mem bers a committee of 5 members is to be formed, then answer the following questions. 3. Sol.

4. Sol.

5. Sol.

Number of committees consisting of 3 official and 2 non-official members, are (A) 363 (B*) 336 (C) 236 (D) 326 8 3 official out of 8 can be selected by C3 = 56 ways 2 non-official out of 4 can be selected in 4C2 = 6 ways  required number of committees are 56 × 6 = 336. Number of committees consisting of at least two non-official members, are (A*) 456 (B) 546 (C) 654 (D) 466 Two non-officials and 3 officials i.e. 4 C2 × 8C3 = 6 × 56 = 336. Three non-official and 2 officials 4 C3 × 8C2 = 4 × 28 = 112. Four non-officials and 1 official 4 C4 × 8C1 = 1 × 8 = 8 Total 336 + 112 + 8 = 456. Number of committees in which a particular official member is never included, are (A) 264 (B) 642 (C) 266 (D*) 462 Required no. of ways = 12 – 1C5 = 11C5 = 462

[P & C, E]

vuqPNsn #1


8ljdkjhvkSj4xSj&ljdkjhlnL;gS]bu12 lnL;ksaesals5 lnL;ksadh,dlfefrcukbZtkrhgSArksfuEufyf[kr

iz'uksadsmÙkjnhft,A 3. Sol.

3ljdkjhvkSj2xSj&ljdkjhlnL;ksadksysdjcukbZtkusokyhlfefr;ksadh la[;kgS& (A) 363 (B*) 336 (C) 236 (D) 326 8 8esals 3ljdkjhlnL;ksadkspquusdsrjhds = C3 =56

[P & C, E]

4esals2xSj&ljdkjhlnL;ksadkspquusadsrjhds =4C2 =6

lfefr cukus ds dqy rjhds = 56 × 6 = 336.

 4.

delsdenksxSj&ljdkjhlnL;ksadksysdjcukbZtkusokyhlfefr;ksadhla[;kgS& (A*) 456

Sol.

(B) 546

(C) 654

(D) 466

nksxSj&ljdkjhrFkk3ljdkjhlnL;ksaokyhlfefr;k¡ = 4C2 × 8C3 = 6 × 56 = 336. rhuxSj&ljdkjhrFkk2ljdkjhlnL;ksaokyhlfefr;k¡ = 4C3 × 8C2 = 4 × 28 = 112. pkjxSj&ljdkjhrFkk1ljdkjhlnL;ksaokyhlfefr;k¡ = 4C4 × 8C1 = 1 × 8 = 8 vr% dqy lfefr;k¡ = 336 + 112 + 8 = 456.

5.

,dfo'ks"kljdkjhlnL;dksdHkhlfEefyrughadjrsgq,cukbZtkusokyh lfefr;ksadhla[;kgS& (A) 264

Sol.

(B) 642 vHkh"V rjhds = 12 – 1C5 = 11C5 = 462

(C) 266

(D*) 462

Comprehenssion # 2 Let n be the number of ways in which the letters of the word "RESONANCE" can be arranged so that vowels appear at the even places and m be the num ber of ways in which "RESONANCE" can be arrange so that letters R, S, O, A, appear in the order same as in the word RESONANCE, then answer the following questions. 6. Sol.

The value of n is (A) 360 (B*) 720 In the word RESONANCE there are 9 letters. Consonants (5), 1R, 1S, 1C and 2N Vowels (4), 2E, 1O, 1A total even places 4 ; No. of ways arranging vowels in even places is

(C) 240

(D) 840

4! = 12 2!

No. of ways arranging consonants in rem aining odd places is

5! 2!

= 60

required number of arrangem ent = 12 × 60 = 720 = n 7.

The value of m is (A*) 3780

(B) 3870

Sol.

Required number of arrangements are

8.

The exponent of 5 in n! is (A) 88 (B*) 178 Now exponents of 5 in 720

Sol.

=

(C) 3670

(D) 3760

9! = 3780 2! 2! 4 !

(C) 358

(D) None of these

 720   720   720   720   5    52    53    54         

= 144 + 28 + 5 + 1 = 178

vuqPNsn #2 PERMUTATION & COMBINATION - 50

ekukfd'kCn"RESONANCE"dsv{kjksadksO;ofLFkrdjusdsmurjhdksadhla[;kgSftuesaLojleLFkkuksaijvkrs gS]ngSavkSj'kCn"RESONANCE"dsv{kjkasdksO;ofLFkrdjusdsmurjhdksadhla[;kgSftuesav{kjR,S,O,A,blh ØeeasvkrsgSa]mgSrksfuEufyf[kriz'uksadsmÙkjnhft,& 6. Sol.

n dk eku gS & (A) 360 (B*) 720 'kCn RESONANCE esadqy9v{kjgSA

(C) 240

(D) 840

O;atu (5), 1R, 1S, 1CvkSj2N Loj (4), 2E, 1O, 1A dqyleLFkkuksadhla[;k=4 LojksadksleLFkkuksaijO;ofLFkrdjusdsrjhdksadhla[;k =

4! =12 2!

'ks"kfo"keLFkkuksaijO;atuksadksO;ofLFkrdjusdsrjhdksdhla[;k= 7.

vr% vHkh"V rjhdks dh la[;k = n = 12 × 60 = 720 m dk eku gS & (A*) 3780

(B) 3870

Sol.

vHkh"V rjhdks dh la[;k = m =

8.

n!esa5dk?kkrkadgS& (A) 88

Sol.

5! =60 2!

(C) 3670

(D) 3760

(C) 358

(D)buesalsdksbZugha

9! =3780 2 ! 2! 4 !

(B*) 178

720esa5dk?kkrkad =

 720   720   720   720   5    52    53    54         

= 144 + 28 + 5 + 1 = 178

PART - III : ASSERTION / REASONING

Hkkx - III : [email protected] (ASSERTION/REASONING) 9.

STATEMENT-1 :

(n  1)! (n  1)!

is divisible by 6 for some n

 N.

[Rank Booster 08-09]

STATEMENT-2 : Product of three consecutive integers is divisible by 3!. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B*) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True

dFku-1 :

(n  1)! (n  1)!

,

dqN n  N ds fy, 6 ls foHkkftr gSA

dFku-2: rhuØekxriw.kk±dksdkxq.kuQy3! lsfoHkkftrgksrkgSA (A) dFku&1 lR; gS] dFku&2 lR; gS ; dFku&2, dFku&1 dk lgh Li"Vhdj.k gSA (B*) dFku&1 lR; gS] dFku&2 lR; gS ; dFku&2, dFku&1 dk lgh Li"Vhdj.k ugha gSA (C) dFku&1 lR; gS] dFku&2 vlR; gSA (D) dFku&1 vlR; gS] dFku&2 lR; gSA


Sol.

Statement-1 :

(n  1)! (n  1)!

= (n + 1) n is divisible by 6 for some n

 N like 2, 3 etc.

[P & C, E]

Statement-2 : Product of three consecutive integers is always divisible by 3!. both are true but statement-2 is NOT a correct explanation for statement-1. Hindi

dFku -1 :

(n  1)! (n  1)!

= (n + 1) n, dqN n  N ds fy, 6 ls foHkkftr gS] tSls 2, 3 ds fy,

[P & C, E]

dFku-2: rhuØekxriw.kk±dksdkxq.kuQylnSo3! lsfoHkkftrgksrkgSA nksuksadFkulR;gSysfdudFku -2,dFku-1dksLi"VughadjrkgSA 10.

Statement -1 : The maximum number of points of intersection of 8 unequal circles is 56. Statement -2 : The maximum number of points into which 4 unequal circles and 4 non coincident straight lines intersect, is 50. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B*) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True [Rank Booster 08-09] dFku-1:8vlekuo`Ùkkssadsvf/kdreizfrPNsnfcUnqvksadhla[;k56gSA

dFku-2:4vlekuo`ÙkksavkSj4vlaikrhljyjs[kkvksadsvf/kdreizfrPNsnfcUnqvksadhla[;k50 gSA (A) dFku&1 lR; gS] dFku&2 lR; gS ; dFku&2, dFku&1 dk lgh Li"Vhdj.k gSA (B*) dFku&1 lR; gS] dFku&2 lR; gS ; dFku&2, dFku&1 dk lgh Li"Vhdj.k ugha gSA (C) dFku&1 lR; gS] dFku&2 vlR; gSA (D) dFku&1 vlR; gS] dFku&2 lR; gSA [Rank Booster 08-09] Sol.

(B) Statement -1: Two circles intersect in 2 points.  Maximum number of points of intersection = 2 × number of selections of two circles from 8 circles. = 2 × 8C2 = 2 × 28 = 56 Statement -2: 4 lines intersect each other in 4C2 = 6 points. 4 circles intersect each other in 2 × 4C2 = 12 points. Further, one lines and one circle intersect in two points. So, 4 lines will intersect four circles in 32 points.  Maximum number of points = 6 + 12 + 32 = 50. Hindi. (B) dFku-1:nkso`Ùkvf/kdre2fcUnqvksaijizfrPNsndjrsgSa



vf/kdreizfrPNsnhfcUnqvksadhla[;k =2×(8o`Ùkksaesals2o`Ùkksadksp;udjusdsrjhds)

= 2 × 8C2 = 2 × 28 = 56 dFku -2: 4js[kk,¡ ijLij4C2 =6fcUnqvksaijizfrPNsndjrhgSA 4 o`ÙkijLij2×4C2 = 12fcUnqvksaijizfrPNsndjrsagSA

 11.

iqu%,djs[kk,oa,do`Ùk]2fcUnqvksaijizfrPNsndjrsgSa vr%4js[kk,¡4o`Ùkksadks32fcUnqvksaijizfrPNsndjsxhA vf/kdre fcUnqvksa dh la[;k =6+12+32 =50.

Statement -1 : If there are six letters L1, L2 , L3, L4, L5, L6 and their corresponding six envelopes E 1, E2, E3, E4, E5, E6. Letters having odd value can be put into odd value envelopes and even value letters can be put into even value envelopes, so that no letter go into the right envelopes, the number of arrangement will be equal to 4. Statement -2 : If Pn number of ways in which n letter can be put in ‘n’ corresponding envelopes such that no

 1 1 ( 1)n    1 – ....  letter goes to correct envelope, then Pn = n!   1!  2!  n !   (A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False PERMUTATION & COMBINATION - 52

(D) Statement-1 is False, Statement-2 is True dFku-1: ;fn 6 i=k L1, L2 , L3, L4, L5, L6 vkSj muds laxr 6 fyQkQs E1, E2, E3, E4, E5, E6gSA fo"ke vadks okys i=k

i=kfo"ke vadokysfyQkQsesaghj[kstkldrs gSarFkklevadksaokysi=klevad okysfyQkQsesaghtkldrsgSa rkfddksbZHkhi=klghfyQkQsesauj[kktk;srcfoU;klksadhla[;k4 gksxhA dFku-2:;fnPn, ni=kksadksnlaxrfyQkQksaesaj[kusdsrjhdksdhla[;kgSrkfddksbZHkhi=klghfyQkQsesau tk;s]rc  1 1 (1)n    1 – ....  Pn = n!   1!  2!  n !   (A*) dFku&1lR; gS] dFku&2 lR; gS ; dFku&2, dFku&1 dk lgh Li"Vhdj.k gSA (B) dFku&1lR; gS] dFku&2 lR; gS ; dFku&2, dFku&1 dk lgh Li"Vhdj.k ugha gSA (C) dFku&1 lR; gS] dFku&2 vlR; gSA Sol.

(D) dFku&1 vlR; gS] dFku&2 lR; gSA (A) L1 L3 L5 L2 L4 L6 E1 E3 E5 E2 E4 E6

 

Number of ways = 3 ! 1  Hindi. (A) L1 L3 L5 E1 E3 E5

 1 1 1   . 3 ! 1     = 4 1! 2 ! 3 !   1! 2! 3 !  1



1



1 

L2 L4 L6 E2 E4 E6

 

1

1

1 

 

1

1

1 

dqy rjhds = 3 ! 1  1!  2!  3 !  . 3 ! 1  1!  2!  3!  = 4 12.





Statement -1 : The maximum value of K such that (50)k divides 100! is 2. Statement -2 : If P is any prime number, then power of P in n! is equal to

n  n n + + P  P 2   P 3  ....      

where [ · ] represents greatest integer function. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D*) Statement-1 is False, Statement-2 is True dFku-1 :Kdk vf/kdre eku tcfd (50)k , 100! dks foHkkftr djsa] 2gSA

n   

n n  +   ....ds cjkcj gksxh] P 2   P3 

dFku-2 : ;fn P,d vHkkT; la[;k gS rc n! esa Pdh ?kkr  P  + 

tgk¡ [·]egÙkeiw.kk±dQyugSA (A) dFku&1 lR; gS] dFku&2 lR; gS ; dFku&2, dFku&1 dk lgh Li"Vhdj.k gSA (B) dFku&1lR; gS] dFku&2 lR; gS ; dFku&2, dFku&1 dk lgh Li"Vhdj.k ugha gSA (C) dFku&1 lR; gS] dFku&2 vlR; gSA (D*) dFku&1vlR; gS] dFku&2 lR; gSA Sol.

(D)

If power of P is l in n! then

l

=

n  n   P  +  P 2  + .....    

Power of 5 in 100! is =

100   5  +  

100   5 2  +  

100   53  = 20 + 4 + 0 = 24  

Power of 2 in 100 ! is =

100  100  100  100  100  100   2  +  2 2  +  23  +  2 4  +  25  +  2 6  = 50 + 25 + 12 + 6 + 3 + 1 = 97            

So power of 50 in 100 ! is 12. PERMUTATION & COMBINATION - 53

Maximum value of K is 12.

n  

n  

;fn n! esa Pdks ?kkr l gks] rks l =  P  +  P 2  +.....

Hindi. (D)

100! esa5dh?kkr =

100  100  100   5  +  5 2  +  53  = 20 + 4 + 0 = 24      

100! esa2dh?kkr =

100  100  100  100  100  100   2  +  2 2  +  23  +  2 4  +  25  +  2 6  = 50 + 25 + 12 + 6 + 3 + 1 = 97            

vr% 100!esa 50dh ?kkr 12gS K dk vf/kdre eku 12 gSA 13.

Statement -1 : If a, b, c are positive integers such that a + b + c  8, then number of possible values of the ordered triplets (a, b, c) is 56 Statement -2 : The number of ways in which n identical things can be distributed into r different groups is n–1Cr–1 (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C*) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True dFku-1 : ;fn a, b, c /kukRed iw.kk±d bl izdkj gS fd a + b + c  8 rks Øfedf=kd (ordered triplets) (a, b, c) dh

laHkkforla[;k56gSA dFku-2:nloZleoLrqvksadksrfofHkUuoxksZesaforfjrdjusdsrjhdsn–1Cr–1gksaxsaA (A) dFku&1 lR; gS] dFku&2 lR; gS ; dFku&2, dFku&1 dk lgh Li"Vhdj.k gSA (B) dFku&1lR; gS] dFku&2 lR; gS ; dFku&2, dFku&1 dk lgh Li"Vhdj.k ugha gSA (C*) dFku&1lR; gS] dFku&2 vlR; gSA (D) dFku&1 vlR; gS] dFku&2 lR; gSA Sol.

a + b+ c = 8 – t 0  t  5 5

 t 0

7 – t

C2 = 56

Ans. (C) 14.

Statement -1 : If N is number of positive integral solutions of x1 x2 x3 x4 = 770, then N is divisible by 4 distinct primes. Statement -2 : Prime numbers are 2, 3, 5, 7, 11, 13, ..... (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D*) Statement-1 is False, Statement-2 is True dFku-1: ;fnx1 x2 x3x4 =770esa/kukRediw.kk±dgyksadhla[;kNgks]rksNpkjfHkUuvHkkT;la[;kvksalsHkkT;gksxk

dFku -2 : vHkkT; la[;k,¡ 2, 3, 5, 7, 11, 13, ..... gSA (A) dFku&1 lR; gS] dFku&2 lR; gS ; dFku&2, dFku&1 dk lgh Li"Vhdj.k gSA (B) dFku&1lR; gS] dFku&2 lR; gS ; dFku&2, dFku&1 dk lgh Li"Vhdj.k ugha gSA (C) dFku&1 lR; gS] dFku&2 vlR; gSA (D*) dFku&1vlR; gS] dFku&2 lR; gSA Sol.

x 1 x 2 x 3 x 4 = 2 × 5 × 7 × 11 N = 44 Ans. (D)


PART - IV : TRUE / FALSE

Hkkx - IV : lR;@vlR; (TRUE / FALSE) 15.

Number of ways in which five vowels of English alphabets and ten decimal digits can be placed in a row such that between any two vowels odd number of digits are placed and both end places are occupied by vowels is 20 (10 !) (5 !)

vaxsztho.kZekykdsik¡pLojksarFkknln'keyovadksadks,diafDresablrjgO;ofLFkrfd;ktk,fdnksLojkasadse/; fo"kela[;kesavadvk;sarFkknksuksafljksaijLojvk;sadsrjhdksadhla[;k20(10!)(5!)gksrhgSA Ans. True (lR;) Sol.

Ten digits can be partitioned into four parts as 1 + 1 + 3 + 5 ; 1 +1 +1+ 7;1 +3+ 3+ 3 (each partitioning has odd number of digits) The number of ways in which these can be placed in the four spaces =

4! 4! 4! + + = 20 ways 2! 3! 3! also numbers of arrangements of vowels = 5 ! Number of arrangements of digits = 10 ! total ways = 20 (10 !) (5 ! ) Hindi.

nlvadksadksblizdkjfoHkkftrfd;ktkldrk gS 1 + 1 + 3 + 5 ; 1 + 1 + 1 + 7 ; 1 + 3 + 3 + 3 ¼izR;sd Hkkx esa fo"ke la[;k esa vad gS½ murjhdksadhla[;kftuesabudkspkjLFkkuksaijj[kktkldrkgS =

4! 4! 4! + + = 20 2! 3! 3!

lkFkghLojksadksO;ofLFkrdjusdsrjhds =5 ! vadksadksO;ofLFkrdjusdsrjhds =10! dqy rjhds = 20 (10 !) (5 !) 16.

In a 12 storey house 10 people enter the lift cabin. It is known that they will leave the lift in predecided groups of 2, 3 and 5 people at different storeis. The number of ways they can do so if the lift does not stop up to the second storey is 720

,d 12eaftyk edku dh fy¶V esa 10O;fDr p<+rs gSa ;g lqfuf'pr dj fn;k x;k gS fd O;fDr fy¶V ls 2,3 rFkk5 dslewgksaesavyx&vyxeaftyijmrjsaxsa]rks;gdk;Zdqy720 rjhdksalsfd;ktkldrkgS;fnfy¶Vnwljheafty rdugha:drhgksA Ans. True (lR;) Sol. Hindi.

The number of ways = 10C3 × 3 ! = 720 rjhdksa dh la[;k = 10C3 × 3 ! = 720

17.

2m white identical coins and 2n red identical coins are arranged in a straight line with (m + n) identical coins on each side of a central mark . The number of ways of arranging the identical coins , so that the arrangements are symmetrical with respect to the central mark, is 2m + 2nC2m

2m loZlelQsnflDdksarFkk 2n loZleykyflDdksdks,ddsUnzh;fpUgdsnksukasrjQ ,d ljyjs[kkesablrjg

O;ofLFkr djrs gSa fd izR;sd rjQ (m + n) loZle flDdsvk;sa] rks O;ofLFkrdjus ds mu rjhdksadh la[;k tcfd O;oLFkkiudsUnzh;fpUgdsnksuksarjQlefergks] 2m+2nC2mgksrhgSaA Ans. False (vlR;) Sol.

 Hindi.

Arrange ments will be one side m+n Cm

,drjQO;ofLFkrgksxsaA


vr% rjhdksa dh la[;k=m+nCm 18.

Three ladies have each brought their one chlid for admission to a school. The principal wants to interview the six persons one by one subject to the condition that no mother is interviewed before her chlid. The number of ways in which interviews can be arranged is 90

rhuefgyk,¡izR;sdvius,dcPps dksfo|ky;esaizos'kfnykusvkrh gSAiz/kkuk/;kidN%O;fDr;ksadk,dds ckn,d lk{kkRdkjysukpkgrkgStcfdfdlhHkhekadkmldscPpslsigyslk{kkRdkjughafy;ktkrkgS]rkslk{kkRdkjysusdsdqy rjhds90gSaA Ans. True (lR;) Sol.

Each lady and her child can be arranged in a fixed order only.



6! 2 ! 2 ! 2 ! = 90

izR;sd efgyk rFkk mldk cPpk ,d fuf'pr Øe esa O;ofLFkr fd;s tk ldrs gSaA

Hindi.

 19.

The total no. of ways in which interview can be held =

6!

vr% lk{kkRdkj ysus ds dqy rjhdksa dh la[;k = 2 ! 2 ! 2 ! = 90

The number of ways one can put three balls numbered 1, 2, 3 in three boxes labelled a, b, c such that at the most one box is empty is equal to 24

rhuxsanksaftuij1,2,3la[;k,¡vafdrgSaArhuckWDlksaesa]ftuija,b,c yscyyxsgq,gSa]24rjhdksalsMkyk tk ldrk gS tcfd vf/kdre 1 ckWDl [kkyh jgsA Ans. True (lR;) Sol.

If none of the boxes is empty. Number of ways = 3! = 6 If one of the boxes is empty The number of ways in which 3 balls can be divided into the groups containing 1 and 2 balls respectively = 3C 1 = 3 Now the number of ways end selection can be put in boxes = 3! = 6  the number of ways in which one box remains empty = 6 × 3 = 18  total number of ways = 6 + 18 = 24

Hindi.

tc dksbZ ckWDl [kkyh u gks] rks rjhdksa dhla[;k =3!=6 ;fn,dckWDl[kkyhgksrks murjhdksadhla[;kftuls3xsanksadksØe'k%1rFkk2xasnksadslewgksaesack¡VktkrkgSA = 3C1 = 3

 

vcbuxsanksadks ckWDlksaesaO;ofLFkrdjusdsrjhds=3!=6 mu rjhdksa dh la[;k tc ,d ckWDl [kkyh jg tk, = 6 × 3 = 18 rjhdksa dh dqy la[;k = 6 + 18 = 24

PART - V : FILL IN THE BLANKS

Hkkx-V : fjDr LFkkuksa dh iwfrZ dhft, (FILL IN THE BLANKS) 20.

Words are formed by arranging the lette rs of the word "STRANGE" in all possible ma nner. Let m be the number of words in which vowels do not come together and 'n' be the num ber of words in which vowels come together. Then the ratio of m: n is_________ .

'kCn"STRANGE" dsv{kjksadkslHkhlEHkorjhdksalsO;ofLFkrdj'kCn cuk;stkrsgSaAekukfd'm'mu'kCnksadh la[;kgSftuesaLoj,dlkFkughavkrsgSavkSj 'n' mu'kCnksadhla[;kgSftuesaLoj,dlkFkvkrsgS]rksvuqikr m: n ________ gSa


Ans. Sol.

21.

5: 2

5 ! 6C2 . 2! 6 ! 2!



5 2

A 5 digit number divisible by 3 is to b e for med using the numerals 0, 1, 2, 3 , 4 & 5 witho ut re petition , then the total number of ways in which this can be done is ________.

vadks 0,1,2,3,4 ,oa 5 dhlgk;rkls 5 vadksdh ______ la[;k,¡cukbZtkldrhgSa]tks 3 lsfoHkkftrgkstcfd vadksdhiqujko`fÙkughagksldrhgSaA Sol.

Ans. 216 When 0 is not being used = 5! When 3 is not being used = 5! – 4 ! Total number of ways = 5 ! + 5 ! – 4 ! = 216

Hindi.

tc 0 dk mi;ksx ugha djsa = 5! tc 3 dk mi;ksx ugha djsa = 5! – 4 ! dqy rjhdksa dh la[;k = 5 ! + 5 ! – 4 ! = 216

22.

There are m points on a straight line AB & n points on the line AC none of them being the point A. Triangles are formed with these points as vertices, when (i) A is excluded

(ii) A may be included. The ratio of number of triangles in the two cases is____.

,dljyjs[kk ABijm fcUnqfLFkrgSrFkknwljhljyjs[kk ACijnfcUnqfLFkrgSaAbuesafdlhesaHkh AlfEefyr ughagSAbufcUnqvksadks'kh"kZekudjf=kHkqtcuk;stkrsgSa]tcfd (i)A 'kkfeyughagS (ii)A'kkfeygksldrkgSaAnksuksafLFkfr;ksaesaf=kHkqtksadhla[;kdkvuqikr _____ gSaA Ans.

m

Sol.

23.

m

m n 2 m n C2 . nC1  m C1 . nC2 n

m

n

C2 . C1  C1 . C2

 m.n



m  n – 2 mn

The number of ways in which the number 94864 can be resolved as a product of two factors is____.

la[;k94864dks ____ rjhdksalsnks xq.ku[k.Mksadsxq.kuQyds:iesaO;Drfd;ktk ldrkgSaA Ans.

23 number of ways rjhdksa dh la[;k=

5.3.3 1 = 23 2

Sol.

94864 = 2 4 . 7 2 . 112

24.

The number of ways in which an insect can move from left bottom corner of a chess board to the right top corner, if it is given that it can move only upside or right, along the lines is ____.

;fn'krjatdscksMZij,ddhM+kjs[kkvksadsvuqfn'kdsoyÅijdhvksj ;knka;hvksjxfrdjldrkgSa]rks;g dhM+kcka;hvksjdsfupysdksusls nka;hvksjdsÅijhdksusrd ____ rjhdksalsxfrdjldrkgSa\ Ans. Sol.

16

C8

( 8  8) ! = 16 C8 818 !


25.

The number of three digit numbers of the form xyz such that x < y and z

 y is____.

xyz :i dh rhu vadks dh ____la[;k,¡ gksxh tcfd x < y vkSj z  y gks Ans.

276

Sol.

= 3

x < y and z  y

= 8

= 15

= 24

= 35

= 48

= 63

= 80 Total = 276

PART - I : IIT-JEE PROBLEMS (PREVIOUS YEARS)

Hkkx-I :IIT-JEEdsiz'u¼iwoZorhZo"kZ½ * Marked Questions are having more than one correct option. * fpfUgr iz'u ,d ls vf/kd lgh fodYi okys iz'u gS IIT-JEE - 2000 1.

How many different nine digit numbers can be formed f rom the number 223355888 by rearranging its digits so that the odd digits occupy even positions? [IIT-JEE-2000, Scr, (1, 0), 35] la[;k223355888dsvadksdksysdj 9vadksdhfdruhla[;k;sacukbZtkldrhgStcfdfo"kevadleLFkkuksaij

vk;sA (A) 16 Sol.

(B) 36

Four digits 3,3,5,5 can be arranged at four even places in be arranged at 5 odd places in

Hindi

(C*) 60

[IIT-JEE-2000,Scr, (1, 0) , 35] (D) 180

4! 2! 2! = 6 ways and remaining digits 2,2,8,8,8 can

5! 2! 3 ! = 10 ways. Number of possible arrangements is 6×10 = 60. 4!

pkjvad3,3,5,5dkspkjleLFkkuksaijj[kusdsrjhds 2! 2! =6gSa]vkSj'ks"kvadks2,2,8,8,8dksik¡pfo"keLFkkuksa PERMUTATION & COMBINATION - 58

5!

ij j[kus ds rjhds 2! 3 ! =10gSaA rc laHkkfor rjhdksa dh la[;k 6×10=60gSA IIT-JEE - 2001 2.

Sol.

Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. If T n+1 – Tn = 21, then n equals [IIT-JEE-2001,Scr, (1, 0) , 35] ekukfd;fnTnmuf=kHkqtksdhla[;kdkscrkrkgStksn–Hkqtkvksaokys,dlecgqHkqtds'kh"kks±dkstksM+uslscursgSa] [IIT-JEE-2001,Scr, (1, 0) , 35] ;fn Tn+1–Tn =21 gks] rks n= (A) 5 (B*) 7 (C) 6 (D) 4 A regular polygo n of n sides has n ver tices num ber two of which a re collinear. so out of n points we have to select 3 sides  T n = nC3 and T n+1 = n+1 C3 According to questions  T n+1 – T n = 2 ! = n+1 C3 – nC3 = 2 ! =

(n  1)n (n – 1) 6

–

n(n – 1) (n – 2) 6

= 21

= n (n – 1) (n + 1 – n + 2) = 21 × 6

21 6 = 42 3 n (n –1) = 7 × 6 n=7 = n (n – 1) =

 3.

Sol.

Let E = {1, 2, 3, 4} and F = {1, 2}. Then the number of onto functions from E to F is [IIT-JEE-2001,Scr, (1, 0), 35] ekuk E = {1, 2, 3, 4} rFkk F = {1, 2} gS] rks E ls F esa vkPNknd (onto) Qyuksa dh la[;k gS& (A*) 14 (B) 16 (C) 12 Here E = (1, 2, 3, 4) and F = (1, 2) Here number of onto function is total number of solution - number of into function = 2 4 – 2 = 16 – 2 = 14

[IIT-JEE-2001,Scr, (1, 0), 35] (D) 8

IIT-JEE - 2002 4.

The number of arrangements of the letters of the word BANANA in which the two N’s do not appear adjacently is

[IIT-JEE-2002,Scr, (3, – 1), 90] 'kCn BANANAdsv{kjksadksysdjfdrus'kCncuk;stkldrsgSatcfdnksuksa N,dlkFkuvk;s – [IIT-JEE-2002,Scr, (3, – 1), 90] (A*) 40 Sol.

(B) 60

(C) 80

Total number of arrangement of word BANANA is

(D) 100

6! = 60 2!. 3 !

and number of ways of two N’s together is

5! 3!

= 20

Here the required number of arrangement if two N’s are not together is 60 – 20 = 40 IIT-JEE - 2004


n  ! 2

5.

Prove by permutation or otherwise

n !n

is an integer (n

 I+).

[IIT-JEE-2004, Main, (2, 0), 60]

n  ! 2

Øep; ds mi;ksx ls ;k vU;Fkk fl) djks fd  n ,d iw.kk±d gS (n I+). n!

[IIT-JEE-2004,Main, (2, 0), 60]

Sol. Here n 2 objects are distributed in n groups, each group containing n identical objects.

 =

=

Number of ways of arrangements n2

2

2

2

Cn .n – n Cn .n – 2n Cn .n – 3n Cn .n (n 2 )!

·

n 2 – n!

n! (n2 – n)! n! (n 2 – 2n)!

...

n! n!. 1



2

– 2n

(n 2 )! (n! )n

Cn ....n Cn

 Integer

(as number of arrangements has to be integer) (n  +) Hindi

;gk¡n2oLrqvksadksnlewgksaesack¡Vkx;kgS]izR;sdlewgesanlekuoLrq,sagSA  O;oLFkkvksadhla[;k =

n2

2

2

2

Cn .n – n Cn .n – 2n Cn .n – 3n Cn .n

(n 2 )!

n 2 – n!

2

– 2n

n! (n 2 )!  · ... = n! (n 2 – n)! n! (n2 – 2n)! n!. 1 (n! )n

Cn ....n Cn

iw.kk±d

(tSlsfdO;oLFkkvksadhla[;k iw.kk±dgksxh)(n+)

IIT-JEE - 2005 6.

A rectangle with sides 2m – 1 and 2n – 1 is divided into squares of unit length by drawing parallel lines as shown in the diagram, then the number of rectangles possible with odd side lengths is [IIT-JEE- 2005,Scr, (3, – 1), 84] ,dvk;rftldhHkqtk,¡2m–1,oa2n–1bdkbZgSdkslekUrjjs[kkvksa}kjk fp=kesafn[kk,vuqlkjbdkbZyEckbZ

dsoxks±esackaVktkrkgS]rksmulEHkovk;rksadhla[;kftudhHkqtkvksadhyEckbZfo"kegSa]gksxh& [IIT-JEE- 2005,Scr, (3, – 1), 84]

Sol.

(A) (m + n – 1) 2 (B) 4m+n–1 (C*) m2 n2 Along hor izontal sid e one unit can be taken = (2m – 1) ways three unit can be taken = (2m – 3) ways five unit can be taken = (2m – 5) ways and so on the number of ways of selecting a side horizontally is

(D) m(m + 1)n(n + 1)


(2m – 1) + (2m – 3) + .....+ 3 + 1 = m 2 Similarly, the number of ways of selecting a side vertically is (2n –1) + (2n – 3) + .......+ 3 + 1 = n 2 Hence required numb er of rectangle is m 2n2

7.

If total number of runs scored in n matches is

 n  1  n+1   (2 – n – 2) where n > 1 and the runs scored in  4 

the k th match are given by k. 2 n+1–k, where 1  k  n, find n

[IIT-JEE-2005,Main, (2, 0), 60]

 n  1  n+1  (2 – n – 2) gks tgk¡ n > 1 rFkk k osa eSp esa cuus okys ju  4 

;fn n eSpksa esa cuus okys dqy ju  k. 2n+1–k gks] Ans. 7

tgk¡ 1  k  n.rks n dk eku Kkr djksA

T k = k.2 n+1–k and S n =

Sol.

n

Now,

Sn



 k.2

 n  1  n1   2 n2  4  n

n 1k

= 2n+1

k .2

k

k 1

k 1

Sn

[IIT-JEE-2005,Main, (2, 0), 60]

n   1  2n1. 2 .1  n  n1   2 2 

(sum of A.G.P.)

 n  1  n1   2  n  2 = 2. [2 n+1 – 2 – n] 4   n 1 = 2 4



 n= 7

Hindi. Tk = k.2n+1–k vkSj Sn =

n

vc

Sn



 k.2

 n  1  n1   2 n2  4  n

n1k

= 2n+1

k

k 1

k 1

Sn

k .2

n   1  2n1. 2.1  n  n1   2 2 

(A.G.P. dk ;ksx)

 n  1  n1   2  n  2 = 2. [2 n+1 – 2 – n] 4  



n 1 = 2 4

 n= 7

IIT-JEE - 2006


8.

Sol.

If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r 2 t4s2, then the number of ordered pair (p, q) is [IIT-JEE-2006, (3, –1), 184] ;fn r,s,tvHkkT; la[;k,¡ gks rFkk p,q/kukRed iw.kk±d bl izdkj gks fd prFkk qdk y-l-i- r 2t4s2 gS] rks Øfer [IIT-JEE-2006, (3, –1), 184] ;qXeksa (p,q)dhla[;kgksxh – (A) 252 (B) 254 (C*) 225 (D) 224 2 4 2 According to quest ion L.C.M. of (p, q) = r t s . For locas (p, q) to contain r 2 following are the possible powers of r in P and Q (0, 2) ; (2 , 2) , (1, 2) out of those all except (2, 2) can be interchanged for p & q. Total ways for p, q to contain different powers of r are = 2 × 3 –1 = 5 Similarly, different powers of t that can be given to P and q are (0, 4), (1, 4), (2, 4), (3, 4) and (4, 4) out of which except (4, 4) all others can be interchanged for p and q Total ways for p, q to contain different powers of t are 2 × 5 – 1 = 9 Similarly total ways for p,q to c ontain different power of s are 2×3 – 1 = 5 Here total required pairs of p,q can be = 5× 9 × 5 = 225

IIT-JEE - 2007 9.

The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an english dictionary. Then number of words that appear before the word COCHIN is [IIT-JEE-2007,P-II, (3, – 1), 81] COCHIN'kCndsv{kjksalsfufeZrlHkh'kCnksadksvaxsth'kCndks"kdsØekuqlkjO;ofLFkrfd;ktk;s]rks'kCn COCHIN

dsØelsiwoZdqyfdrus'kCngSA (A) 360 Sol.

(B) 192

COCHIN 

C-2 H-1 I-1

Number of words

(C*) 96

(D) 48 [IIT-JEE-20 07, P-II, (3, – 1), 81]

N-1 O-1 = 24

= 24

= 24

= 24 _____ 96

Before word of cochin is the number of words = 96 IIT-JEE - 2008 10.

Consider all possible permutations of the letters of the word ENDEANOEL. Match the Statements/ Expressions in Column I with the Statements / Expressions in Column II and indicate your answer by darkening the appropriate bubbles in the 4 × 4 matrix given in the ORS. [IIT-JEE-20 08,P-II, (6, 0), 81]

(A) (B)

Column I

Column II

The number of permutations containing the word ENDEA is The number of permutations in which the letter E occurs in the first and the last positions is

(p) (q)

5! 2 × 5!


(C)

The number of permutations in which none of the letters D, L, N occurs (r) 7 × 5! in the last five positions is (D) The number of permutations in which the letters A, E, O occur only (s) 21 × 5! in odd positions is Ans. (A)  (p), (B)  (s), (C)  (q), (D)  (q) 'kCnENDEANOELdsv{kjksadslHkhlEHkkforØep;yhft;sColumnIesafn;sx;sizdFku@O;atdksadkdkWyeColumn II

esafn;sx;sdFkuksalslqesydjsavkSjviukmÙkj 4×4eSfVªDldsmfprcqYyksa (bubbles)dksdkykdjdsn'kkZ,a [IIT-JEE-2 008, P-II, (6, 0), 81]

dkWyeI (A) (B) (C) (D) Ans. Solution : (A)

(B)

dkWyeII

'kCnENDEAdkslfEefyrdjusokysØep;ksadhla[;kgS izFkevkSjvafrenksuksLFkkuksaijv{kj EokysØep;ksadhla[;kgS vafreik¡pLFkkuksaijD,LvkSjNesalsdksbZv{kjuvkusokysØep;ksa dhla[;kgS dsoyfo"keLFkkuksaijghv{kj A,E,OvkusokysØep;ksadhla[;kgS (A)  (q),

(B)  (s),

Hint :

(D)

5!

(q)

2 × 5!

(r)

7 × 5!

(s)

21 × 5!

(D)  (q)

Consider 'ENDEA' as one letter so there are five things which are ENDEA , N, O, E, L  number of permutations = 5! Hint : Number of arrangements of n objects in a line in which r objects always remain together in a particular order are (n – r + 1)! E N D E A N O L E (first and last letters are f'ixed) Number of permutations =

(C)

(C)  (q),

(p)

7! = 21 × 5! 2!

Number of permutations of n objects in which r are alike =

n! r !

N, N, D, L will come at first four places and E, E, E, O, A will come at last five places 5! 4! × = 2 × 5! 3! 2!



number of permutations =

Hint :


n! r !

A, E, E, E, O, occur at odd positions and N, N, D, L at even positions = Hint :


5! 4! × = 2 × 5! 3! 2!

n! r !

Hindi : (A)

'ENDEA' dks ,d v{kj ekuus ij] vc ;gk¡ ik¡p v{kj ENDEA , N, O, E, L gSA

Øep;ksa dh la[;k=5! Hint : noLrqvksadks,djs[kkesaO;ofLFkrdjusdsrjhdsftuesaroLrq,sages'kk,dfuf'prØeesalkFk jgrhgSdhla[;k (n–r+1)!gksrhgSA E N D EAN O LE (izFke rFkk vfUre v{kj fu;r gSA) 

(B)

Øep;ksa dh la[;k = Hint : (C)

7! = 21 × 5! 2!

noLrqvksaftuesar,dlekugS]dsØep;ksadhla[;k =

n! r !

N,N,D,LigyspkjLFkkuksaesavk;saxsrFkkE,E, E,O,AvfUreik¡pLFkkuksaesavk,saxsA

4! 5! × = 2 × 5! 2! 3!



Øep;ksa dh la[;k =

Hint :


n! r !


(D)

A,E, E,E,Ofo"ke LFkkuksa ij vkrs gSa rFkk N,N,D,L le LFkkuksa ij vkrs gSa = Hint :


5! 4! × =2×5! 3! 2!

n! r !

IIT-JEE 2009 11.

Sol.

The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is [IIT-JEE-20 09, Paper-I, (3, – 1), 240] lkr&vadh;iw.kkZdksadhla[;k]ftudsvadksdk;ksx10gSvkSjtksdsoy1,2o3vadksalscuhgksa]oggS (A) 55 (B) 66 (C*) 77 (D) 88 There are two possible cases Case 1 : Five 1’s, one 2’s, one 3’s Number of numbers =

7!

= 42 5! Case 2 : Four 1’s, three 2’s Number of numbers =

7! = 35 4! 3!

Total number of numbers = 42 + 35 = 77 Hindi.

nksfLFkfr;kalaHkogSA fLFkfr1 : ikap ckj 1,,d ckj 2, ,d ckj 3 la[;kvksadhla[;k =

7! 5!

=42

fLFkfr2 : pkj ckj 1, rhu ckj 2

la[;kvksadhla[;k=

7! =35 4! 3!

dqy la[;kvksa dh la[;k=42+35=77 IIT-JEE 2010 12.

Sol.

Let S = {1, 2, 3, 4}. The total number of unordered pairs of disjoint subsets of S is equal to ekukfd S={1,2,3,4} ] rks Sds vØfer rFkk vla;qDr leqPp; ;qXeksa dh dqy la[;k fuEu gS (A) 25 (B) 34 (C) 42 (D*) 41 [IIT-JEE-2010, Paper-2, (5, –2), 79] S = {1, 2, 3, 4} Each element can be put in 3 ways either in subsets or we don’t put in any subset. So total number of unordered pairs =

Hindi

3 3 3 3 1 + 1 = 41. [Both subsets can be empty also] 2

S = {1, 2, 3, 4}

fdlh mileqPp; esa izR;sd vo;o dks 3rjhdslsj[kktkldrkgS;kfdlhmileqPp;esaughaj[kktkldrkgSA blfy,vØfer;qXeksadhdqyla[;k =

3 33  3 1 +1=41.[nksuksamileqPp;[kkyhHkhgksldrsgS] 2

PART - II : AIEEE PROBLEMS (PREVIOUS YEARS)

Hkkx-II :AIEEEdsiz'u¼iwoZorhZo"kZ½ 1.

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is[AIEEE 2003]

,dfo|kFkhZijh{kkesa13esals10iz'uksadstokcblrjgnsrkgSfdizFkeik¡piz'uksaesalsdelsdepkjds tokc nsukt:jhgSrksmldstokcnsusdsrjhdksadhla[;kgS& PERMUTATION & COMBINATION - 64

&& Sol.

(1) 140 5 C4 . 8C6 + 5C5 . 8C 5 = 140 + 56 = 196

(2*) 196

(3) 280

(4) 346

2.

The number of ways in which 6 men and 5 women can dine at a round table, if no two women are to sit together, is given by [AIEEE 2003]

,dxksyestijN%iq:"krFkk 5efgyk,¡fdrusrjhdksalscSBldrsgS;fndksbZHkhnksefgyk,¡lkFkughacSBldrh gks\ Sol. 3.

(1*) 6! × 5! 6! 5!

(2) 30

(3) 5! × 4!

(4) 7! × 5!

How many ways are there to arrange the letters in the word GARDEN with the vowels in alphabetical or der ? [AIEEE 2004]

'kCnGARDENdsv{kjksadksfdrusrjhdksalsO;ofLFkrfd;ktkldrkgSftlesaLojo.kZekykdsØeesagS\ (1) 120 Sol.

4.

(2) 240

(3*) 360

(4) 480

6! = 360 2! The number of ways of distributing 8 identical balls in 3 distinct boxes, so that none of the boxes is empty, is [AIEEE 2004] 8lekuxsanksdks3lekuckWDlesafdrusrjhdksalsforfjrfd;k tkldrkgStcfddksbZckWDl[kkyhuk jgs\ (1) 5 (2*) 21 (3) 38 (4) 8C3

Sol.

(5  2)! = 21 5 ! 2! 5.

If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial num ber [AIEEE 2005]

;fnSACHIN'kCndsv{kjksalscuusokyslHkhlEHkkfor'kCnksadksvaxzsth'kCndks'kdsvuqlkjØeesafy[kktk,srks bu'kCnksaesaSACHIN'kCndkØekadgksxk& Sol.

(1) 602 ACHINS A .. ....5 ! C ......5 !

(2) 603

(3) 600

(4*) 601

H ......5 ! I ......5 ! N ......5 ! SACHIN 5 . 5 ! + 1 = 601

6.

The set S : = {1, 2, 3 ..........12} is to be partitioned into three s ets A, B, C of equal size. Thus, A  B  C = S, A  B = B  C = A  C =  The number of ways to partition S is -

[AIEEE 2007]

leqPPk; S:= {1,2,3..........12}dks rhu cjkcj leqPp;ksa A,B,oCesa foHkkftr djuk gS bl izdkj dh A  B  C = S, A  B = B  C = A

 C = 

rksSdksfoHkkftrdjusdsrjhdksadhla[;kgS& (1) 12!/3! (4!)3

(2) 12!/3!(3!)4

(3*) 12!/(4!)3

(4) 12!/(3!)4 PERMUTATION & COMBINATION - 65

Permutation Reso Sheet

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