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STUDY PACKAGE Subject : Mathematics Topic : Permutation Per mutation and Combination Combinatio n
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Permutation and Combination
Permutations are arrangements and combinations are selections. In this chapter we discuss the methods of counting of arrangements and selections. The basic results and form ulas are as follows: 1. Fu n da m e n t al P rin ci p le o f Co u nt in g : m ( i) P rin ci p le o f M u lt ip l ic a t io n: If an event can occur in ‘m’ different ways, following which 0 2 o another event can occur in ‘n’ different ways, then total number of dif ferent ways of simultaneous f o c occurrence of both the events in a definite order is m n. 2 . ( ii ) P rin c i p le of Ad di t io n :If an event can occur in ‘m’ diff erent ways, and another event can occur e g g in ‘n’ diff erent ways, then exactly one of the ev ents can happen in m + n ways. a a p Example # 1 There are 8 buses running from Kota t o Jaipur and 10 buses running f rom Jaipur to Delhi. In h how many ways a person can travel f rom Kota to D elhi via J aipur by bus. u Let E1 be the event of travelling from Kota to Jaipur & E2 be the event of t ravelling from Jaipur to . SSolution. 9 y Delhi by the person. 5 5 E1 can happen in 8 ways and E2 can happen in 10 ways. B s Since both the events E1 and E2 are to be happened in order, simultaneously, the number of ways = 8 0 6 h = 80. 2 t Example× 10 # 2 How many numbers between 10 and 10,000 can be form ed by using the digits 1, 2, 3, 4, 5 if 9 a 0 (i) No digit is repeated in any number. (ii) Digits can be repeated. 0 M 9 . Solution. r (i) Num ber of two di git num be rs = 5 × 4 = 20 w e Number of three digit numbers = 5 × 4 × 3 = 60 b w Number of four digit numbers = 5 × 4 × 3 × 2 = 120 Total = 200 m u w (ii) Number of two digit numbers = 5 × 5 = 25 N Number of three digit numbers = 5 × 5 × 5 = 125 & p Number of four digit numbers = 5 × 5 × 5 × 5 = 625 Total = 775 p A mSelf Practice Problems : s t o1. How many 4 digit number s are there, without repetition of digit s, if each number is divi sible by 5. a c Ans. 952 h . 2. Using 6 different f lags, how many different signals can be made by using atleast three flags, arranging W s , one above the other. Ans. 1920 e s2 . Arra n g e m e n t : If nP r denotes the number of permutations of n diff erent things, taking r at a time, then 1 8 s 8 n! a n 8 Pr = n (n 1) (n 2)..... (n r + 1) = l 5 ( n r )! CNOTE : (i) factorials of negative integers are not defined. 0 (ii) 0 ! = 1 ! = 1 ; 3 o(iii) nP = n ! = n. (n 1) ! n 9 (iv) (2n) ! = 2 . n ! [ 1. 3. 5. 7... (2n 1)] k n 9 eExample # 3: How many numbers of three digits can be f ormed using the digits 1, 2, 3, 4, 5, without repetition 8 of digits. How many of these are even. T 0 . Solution.: Three places are to be filled with 5 diff erent objects. , w 0 Number of ways = 5P 3 = 5 × 4 × 3 = 60 0 w For the 2nd part, unit digit can be fil led in two ways & the remaining two digits can be fill ed in 4P2 ways. 0 4 w Number of even numbers = 2 × P 2 = 24. 0 Example # 4: If all the letters of the word 'QUEST' are arranged in all possible ways and put in dictionary 0 : 2 order, then find the rank of the given word. 3 eSolution.: 4 t Number of words beginning with E = P = 24 ) i 4 5 3 Number of wards beginning with QE = P 3 = 6 s 5 7 Number of words beginning with QS = 6 b 0 Number of words beginning withQT = 6. ( e : Next word is 'QUEST' its rank is 24 + 6 + 6 + 6 + 1 = 43. wSelf Practice e Problems : n Find the sum of all f our digit numbers (without repetition of dig its) form ed using the digits 1, 2, 3, 4, 5. o m3. h o Ans . 399960 P r n–1 n l Find 'n', if P3 : P4 = 1 : 9. Ans. 9 f 4. . a 5. Six horses take part in a race. In how many ways can these horses come in the f irst, second and third e p place, if a part icular horse is among the three winners (Assume No Ties). Ans. 60 o g3 . The number of circular permutations of n diff erent things taken all at a h Cir c u la r P e r m u t a ti o n : a time is; (n 1) !. If clockwise & anticlockwise circular permutations are considered to be same, B k , ) c r i (n 1)! a S then it is . . 2 PNote: Number of circular K permutations of n things when p alike and the rest different taken all at a tim e . y
R d (n 1)! . u distinguishing clockwise and anticlockwise arrangement is . S t ( a SExample # 5: In how many ways can we arrange 6 dif ferent f lowers inp!a circle. In how many ways we can form i y r a garland using these fl owers. d a The number of circular arrangements of 6 dif ferent flowers = (6 – 1)! = 120 aSolution.: K When we form a garland, clock wise and anticlockwise arrangements are similar. Therefore, the number . o l R n 1 g of ways of forming garland = (6 – 1) ! = 60. a w 2 oExample # 6: In how many ways 6 persons can sit at a round table, if two of t hem prefer to sit together. h u S DSolution.: Let P , P , P , P , P , P be the persons, where P , P want to sit together. : Regard these person as 5 objects. T hey can be arranged in a circle i n (5 – 1)! = 24. Now P P can be s E h arranged in 2! ways. Thus the tota l number of ways = 24 × 2 = 48. t ESelf Practi ce Problems : 6. In how many ways the letters of the word 'MONDAY' can be writ ten around a a R M circle if t he vowels are to be separated in any arrangement. Ans. 72 , F7. In how many ways we can form a garland using 3 different r ed flowers, 5 diffe rent yellow flowers and 4 s 1
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different blue fl owers, if flowers of same colour must be together. Ans. 17280. Se le ct ion : If n Cr denotes the number of combinations of n diff erent things taken r at a tim e, then n n! Pr n Cr = = where r n ; n N and r W.. r ! (n r )! r ! NOTE : (i) nCr = n Cn – r (ii) nC r + n Cr – 1 = n + 1C r (iii) n Cr = 0 if r {0, 1, 2, 3........, n} Example # 7 Fifteen players are selected for a cricket match.
4.
e s s a l C o k e T
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com (i) In how many ways the playing 11 can be selected
(ii) (iii) Solution.
In how many ways the playing 11 can be selected including a particular player. In how many ways the playing 11 can be selected excluding two particular players. (i) 11 pl ay er s ar e t o be sel ec te d f ro m 15 Number of ways = 15C11 = 1365. (ii) Since one player is already included, we have to select 10 from the remaining 14 0 m 2 Number of ways = 14C10 = 1001. f o o (iii) Since two players are to be excluded, we have to select 11 from the remaining 13. c 13 3 . Number of ways = C11 = 78. e gExample # 8 If 49C3r – 2 = 49C2r + 1, find 'r'. g n n aSolution. a Cr = Cs if either r = s or r + s = n. p h Thus 3r – 2 = 2r + 1 r= 3 u or 3r – 2 + 2r + 1 = 49 5r – 1 = 49 r = 10 . S r = 3, 10 yExample # 9 A regul ar pol ygon has 20 sides. Ho w many t riangles can b e drawn by usin g the v ertices, but 9 5 5 B not using the sides. 0 sSolution. The first v ertex can be selected in 20 ways. The remaining two are to be selected from 17 6 h 17 2 t vertices so that they are not consecutive. This can be done in C2 – 16 ways. 17 9 a The total num ber of ways = 20 × ( C2 – 16) 0 But in this m ethod, each selection is repeated thrice. 0 M
9 . r 20 (17 C 2 16) w e Number of tri angles = = 800. b w 3 u wExample # 10 10 persons are sitting in a row. In how many ways we can select three of them if adjacent m persons are not selected. N &Solution. p Let P , P , P , P , P , P , P , P , P , P be the persons sitting in this order. p If thr ee are selected (non consecutive) then 7 are left out. A m Let PPPPPPP be the lef t out & q, q, q be the select ed. The number of ways in which these 3 q's can t s o a be placed into the 8 positions between the P's (including extremes) is the number ways of required c h . selection. Thus num ber of way s = C = 56. sExample # 11 In how many ways we can select 4 letters from the letters of the word M SS SS PP. W , eSolution. M 1 s 8 s 8 SSSS a 8 l PP 5 C 0 Number of ways of selecting 4 alike letters = C = 2. 3 o Number of ways of selecting 3 alike and 1 dif ferent letters = C × C = 6 9 k 8 Number of ways of selecting 2 alike and 2 alike letters = C = 3 9 e Number of ways of selecting 2 alike & 2 diff erent = C × C = 9 T 0 . Number of ways of selecting 4 diff erent = C = 1 Total = 21 , wSelf Practice Probl ems :8. In how many ways 7 persons can be selected from among 5 Indian, 4 British & 0 0 2 Chinese, if at least two are to be selected from ea ch country. Ans. 100 w9. 0 10 points lie in a plane, of which 4 poin ts are collinear. Barring these 4 points no three of th e 10 points w 0 are collinear. How many quadrilaterals can be dr awn. Ans. 185. : 10. In how many ways 5 boys & 5 girls can sit at a round table so that girls & boys sit alternate. An s. 2880 0 3 e11. In how many ways 4 persons can occupy 10 chairs in a row, if no t wo sit on adjacent chairs. An s. 840. 2 t ) i 12. In how many ways we can select 3 letters of t he word PROPORTION. Ans. 36 5 s5 . The number of permutations of 'n' things, taken all at a t ime, when 'p' of them are simil ar & of one type, 5 b q of them are similar & of another type, 'r' of them are similar & of a third type & the remaining 7 0 ( e : n! w e n (p + q + r) are all diff erent is . n p ! q ! r ! o mExample # 12 In how many ways we can arrange 3 red flowers, 4 yellow flo wers and 5 white f lowers in a row. h o r In how many ways this is possible if the white flowers are to be separated in any arrangement (Flowers P l f . of same colour are identical). a eSolution. p Total we have 12 f lowers 3 red, 4 yel low and 5 white. o g h 12 ! a B Number of arrangements = = 27720. k , 3! 4! 5! ) c r i For the second part, first arrange 3 red & 4 yellow a S . P 7! K . This can be done in = 35 ways y 3! 4! R d . Now select 5 places from among 8 places (including extr emes) & put the white flowers there. u S t ( This can be done in C = 56. a S The number of ways for the 2 part = 35 × 56 = 1960. i r dExample # 13 In how many ways the letters of the word "ARRANGE" can be arranged without altering the y a relativ e positions of vowels & consonants. a K o . 4! l R nSolution. The consonants in their positions can be arranged in 2 ! = 12 ways. g a w h o 3! u The vowels in their positions can be arranged in = 3 ways S D 2! : s Total number of arrangements = 12 × 3 = 26 E t ESelf Practice Probl ems : 13. How many words can be forme d using the letters of th e word ASSESSMENT if h a word begin wit h A and end with T. Ans . 840 R14. each M If al l the lett ers of the word ARRANGE are arranged in all possible ways, in how many of words we will , F s have the A's not together and also the R's not togeth er. Ans. 660 1
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15. 6.
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How many arrangements can be made by taking four letters of the word MISSISSIPPI . Ans. 176. s s Fo rm a t ion o f Gro u p s : Number of ways in which (m + n + p) diff erent things can be divided into three l a C m n p ! diff erent groups containing m, n & p things respectively is , o k m! n! p!
e T
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com If m = n = p and the groups have identical qualitative characteristic then the number of groups =
(3n)!
n! n! n! 3!
However, if 3n things are to be div ided equally among three people then the number of ways =
(3n )!
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n! m oEx.14. 12 different toys are to be distributed to three children equally. In how many ways this can be done. c . Solution. The problem is to divide 12 different thi ngs into three different groups. e g 12 ! g Number of ways = = 34650. a a 4 ! 4 ! 4 ! p h uExample # 15 In how many ways 10 persons can be div ided into 5 pair s. We hav e each group having 2 persons and the qualitative characteristic are same (Since there . SSolution. 9 is no purpose mentioned or names for each pair) . y 5 5 B 10 ! 0 s Thus the number of ways = = 945. 5 6 h ( 2 ! ) 5 ! t Self Practice Pro blems : 16. 9 persons enter a lift from ground fl oor of a building which stops in 10 floors 2 9 a (excluding ground floor). If is known that persons will l eave the lift i n groups of 2, 3, & 4 in dif ferent 0 0 M 9 floor s. In how many ways this can happen. Ans . 907200 . r w e 52 ! b w17. In how many ways one can make f our equal heaps using a pack of 52 playi ng cards. An s. (13 ! )4 4 ! u w18. In how many ways 11 diff erent books can be parcelled into four packets so that three of the packets contain m N & p 11 ! p 3 books each and one of 2 books, if all packets have the same destination. An s. 4 A m (3 ! ) 2 s t o7 . Se le ct io n o f o n e o r m o re ob je c t s a c h . (a) Number of ways in which atleast one object be selected out of 'n' distinct objects is s W C + C + C +...............+ C = 2 – 1 e (b) Number of ways in which atleast one object may be selected out of 'p' alike objects of one type , 1 s 'q' alike objects of second type and 'r' alike of third type is 8 s 8 (p + 1) (q + 1) (r + 1) – 1 a l (c) Number of ways in which atleast one object may be selected from 'n' objects where 'p' alike of 8 5 one type 'q' alike of second type and 'r' alike of third type and rest C 0 3 o n – (p + q + r) are different, is 9 k (p + 1) (q + 1) (r + 1) 2 – 1 9 eExample # 16 There are 12 diffe rent books on a shelf. I n how many ways we can select atleast one of them. 8 T 0 Solution. We m ay select 1 book, 2 books,........, 12 books. . , number of ways = C + C + ....... + C = 2 – 1. = 4095 wExample # 17 The 0 There are 12 fruit s in a basket of which 5 ar e apples, 4 mangoes and 3 bananas (fruits of same 0 w 0 species are identical). How many ways are there to select atleast one frui t. wSolution. 0 Let x be the number of apples being selected 0 y be the number of m angoes being selected and : 2 3 z be the number of bananas being selected. e t ) i Then x = 0 , 1 , 2 , 3 , 4, 5 5 s 5 y = 0, 1, 2, 3, 4 7 b z = 0, 1, 2, 3 0 ( e Total number of triplets (x, y, z) is 6 × 5 × 4 = 120 : w Exclude (0, 0, 0) Number of combinations = 120 – 1 = 119. e n Self Practi ce Problems o m19. In a shelf there are 5 physics, 4 chemistry and 3 mathematics books. How many combinations are h o P there if (i) books of same subject are different (ii) books of same subject are identical. r l f . Ans. (i) 4095 (ii) 119 a e p o g20. From 5 apples, 4 mangoes & 3 bananas in how many ways we can select atleast two fruits of each h a variety if (i ) fruits of same species are identical (ii) fruits of same species are different. B k , Ans. (i) 24 (ii) 1144 ) c8 . r i Coefficient of x in expansion of (1 x) = C (n N) M ult in o m ia l Th e o r e m : a Number of ways in which it is possible to make a selection f rom m + n + p = N things, where p are alike S . P of one kind, m alike of second kind & n alike of t hird kind taken r at a time is given by coefficient of K . y x in the expansion of R d (1 + x + x +...... + x ) (1 + x + x +...... + x ) (1 + x + x +...... + x ). . u (i) For example the number of ways in which a selection of four letters can be made from the S t ( letters of t he word PROPORTION is given by coefficie nt of x in a S y (1 + x + x + x ) (1 + x + x ) (1 + x + x ) (1 + x) (1 + x) (1 + x). i r d (ii) Method of fictious partition : a a K Number of ways in which n identical thi ngs may be distributed among p persons if each person o . l may receive none, one or more things is; C. R nExample # 18: Find the number of solutions of the equation x + y + z = 6, where x, y, z W. g a wSolution. Number of solutions = coefficient of x in (1 + x + x + ....... x ) h o = coefficient of x in (1 – x ) (1 – x) u = coefficient of x in (1 – x) S D : 3 6 1 s E = C = 28. h = t E 6 a RExample # 19: In a bakery four types of biscuits are available. In how many ways a person can buy 10 M , F biscuits if he decide to take atleast one biscuit of e ach vari ety. s n
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Solution. Let x be the number of biscuits the person select from f irst variety, y from the second, z f rom e s the third and w from the fourt h variety. Then the number of ways = number of solutions of the equation s a l x + y + z + w = 10. C where x = 1, 2, .........,7 o y = 1, 2, .........,7 k e z = 1, 2, .........,7 T w = 1, 2, .........,7
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com This is equal to = coefficient of x 10 in (x + x2 + ...... + x7)4 = coefficient of x 6 in (1 + x + ....... + x6 )4 = coefficient of x 6 in (1 – x 7 )4 (1 – x) –4
4 6 1 = 84. 6
=
= coefficient x 6 in (1 – x) –4
mSelf Practice Probl ems: o21. Three distingui shable dice are rolled. In how many ways we can get a total 15. Ans. 10. c . how many ways we can give 5 apples, 4 m angoes and 3 oranges (fruit s of same species are simil ar) e g22. In to three persons if each may receiv e none, one or more. Ans. 3150 g a9 . a Let N = p q r ..... where p, q, r...... are distinct primes & a, b, c..... are natural numbers then : p h (a) The total numbers of divisors of N including 1 & N is = (a + 1) (b + 1) (c + 1)........ u (b) The sum of these divi sors is = . S 9 (p + p + p +.... + p ) (q + q + q +.... + q ) (r + r + r +.... + r )........ y 5 (c) Number of ways in which N can be resolved as a product of two f actors is 5 B 1 (a 1)(b 1)(c 1).... 0 s if N is not a perfect square 6 2 h = 2 t 1 (a 1)(b 1)(c 1)....1 if N is a perfect square 9 a 2 0 (d) Number of ways in which a composite number N can be resolved into two factors which are 0 M . relatively prime (or coprime) to each other is equal to 2 where n is the number of diff erent 9 r w prime factors in N. e b wExample # 20 Find the number of div isors of 1350. Also find the sum of all di visors. m 1350 = 2 × 3 × 5 u wSolution. Number of div isors = (1+ 1) (3 + 1) (2 + 1) = 24 N & sum of div isors = (1 + 2) (1 + 3 + 3 + 3 ) (1 + 5 + 5 ) = 3720. p p Example # 21 In how many ways 8100 can be resolved into pr oduct of two fact ors. A mSolution. 8100 = 2 × 3 × 5 s t o a 1 c h . Number of ways = ((2 + 1) (4 + 1) (2 + 1) + 1) = 23 s 2 W , eSelf Practice Problems : 1 s23. How many divisors of 9000 are even but not divisibl e by 4. Also fi nd the sum of all such div isors. 8 s Ans. 12, 4056. 8 a 8 l 24. In how many ways the number 8100 can be writt en as product of two coprime f actors. An s. 4 5 C1 0 . Let there be 'n' types of objects, with each type cont aining atleast r objects. Then t he number of ways 0 3 of arrangi ng r objects in a row is n . o kExample # 22 How many 3 digit numbers can be formed by using the digits 0, 1, 2, 3, 4, 5. In how many of 9 8 9 e these we have atleast one digit repeated. T 0 Solution. We hav e to fi ll three places using 6 objects (repeatation allowed), 0 cannot be at 100 place. . , w 0 The number of numbers = 180. 0 w 0 Number of numb ers in which no digit is repeated = 100 w 0 Number of num bers in which atleast one digit is repeate d = 180 – 100 = 80 : Example # 23 How many functions can be defined from a set A containing 5 elements to a set B having 3 0 2 3 e elements. How many these are surjectiv e functions. t ) i Solution. Image of each element of A can be taken in 3 ways. 5 s 5 Number of functions from A to B = 3 = 243. 7 b Number of into functions from A to B = 2 + 2 + 2 – 3 = 93. 0 ( e Number of onto functio ns = 150. wSelf Practice Pro blems : 25. Find the sum of all three digit numbers those can be formed by using the : e n digits. 0, 1, 2, 3, 4. o m h Ans. 27200. o r 26. How many functio ns can be defined from a set A containing 4 elements to a set B containing 5 elem ents. P l f . Ans . How many of these are injective functi ons. 625, 120 a e27. In how many ways 5 persons can enter into a auditorium hav ing 4 entries. Ans. 1024. p o g1 1 . D e a r r a n g e m e n t : h a Number of ways in which 'n' letters can be put in 'n' c orresponding envelopes such that no letter go es B k , to correct envelope is ) c r i a S 1 1 1 1 1 . .......... .. ( 1)n n! 1 P K 1 ! 2 ! 3 ! 4 ! n ! . y dExample # 24 In how many ways we can put 5 writin gs into 5 corresponding envelopes so that no writi ng go R . to the corresponding envelope. u S t ( The problem is the number of dearragements of 5 digits. a SSolution. y i 1 1 1 1 r d = 44. This is equal to 5! a a 2 ! 3 ! 4 ! 5 ! K oExample # 25 Four slip of papers with the numbers 1, 2, 3, 4 written on t hem are put in a box. They are drawn . l R n one by one (without replacement) at random. In how many ways it can happen that the ordinal number g a w of atleast one slip coincide with its own number. h oSolution. Total number o f ways = 4 ! = 24. u The number of ways in which ordinal number of any slip does not coincide with its own number is the S D : s 1 1 1 E h = 9 t number of dearrangements of 4 objects = 4 ! E a 2 ! 3 ! 4 ! R M Thus the required numb er of ways. = 24 – 9 = 15 , FSelf Practice Probl ems: s a.
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In a match column question, Column contain 10 questions and Column I I contain 10 answers written e s in some arbitr ary order. In how many ways a student can answer this question so that exactl y 6 of hi s s a l matchings are correct. Ans . 1890 C In how many ways we can put 5 letters into 5 corresponding env elopes so that atleast one letter go t o o k wrong envelope. Ans . 119
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