Permutation and Combination C ombination : Divisibility Question Question 1 How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeating (1) (2) (3) (4) (5)
15 96 216 120 625
Correct Answer is 216 - (3 )
Exp lanation T est
of
Th e
sum
For Sum As Th ere
sh ould Th e
if
of
of
t he
divisibility
digits
of
any
instance, its digits is 21 is divisible
number take 5 + by
t hat
is
for divisible
the + 3 54372
4 '3',
by
'3'
3 is
divisible
number 7 + also
+ is
by
2 = divisible
3.
54372. 21. by
are si x digits viz., 0, 1, 2, 3, 4 and 5. T o form 5-digit numbers we need e x actly 5 digits. So we not be using one of t he digits.
sum of all t he si x digits 0, 1, 2, 3, 4 and 5 is 15. We know th at any num ber is divisible divisible by 3 if and only t he sum of its digits are divisible by '3'.
Combining t he two criteria t hat we use only 5 of t he 6 digits and pick t hem in such a way t hat t he sum is divisible by 3, we s hould not use eit her '0' or '3' w h ile forming t h e five digit numbers. Case 1 If we do not use '0', then t he remaining 5 digits can be arranged in 5! ways = 120 numbers. Case 2 If we do not use '3', th en th e arrangements sh ould take into account that '0' cannot be t he first digit as a 5-digit number will not start wit h '0'. .
Th e
Th en
So,
first
digit
from
t he
left
can
be
any
of
the
4
digits
1,
2,
4
or
5.
t he remaining 4 digits including '0' can be arranged in t h e ot her 4 places in 4! ways. t here
will
be
4*4!
numbers
=
4*24
=
96
numbers.
Combining Case 1 and Case 2, t here are a total of 120 + 96 = 216 5 digit numbers divisible by '3' that can be formed using t he digits 0 to 5.
Question 2 In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys? (1) (2) (3) (4) (5)
20 30 25 600 480
Correct Answer is 25 - C hoice (3)
Exp lanation
3.
Th e
toys
are
different;
Th e
bo x es
are
identical
If none of t he bo x es is to remain em p ty, t hen we can p ack the to ys in one of t he following ways a. b.
2, 3,
Case T wo
a. Number
of
2, 1,
ways
of
achieving
the
first
1 1 o p tion
2
-
2
-
1
toys out of t he 5 can be selected in 5C2 ways. Anot h er 2 out of th e remaining 3 can be selected in ways and t he last toy can be selected in 1C1 way.
3C2
However, as the bo x es are identical, t he two different ways of selecting w hic h bo x h olds t he first two toys and w h ic h one holds t he second set of two toys will look t he same. Hence, we need to divide t he result by 2.
Th erefore,
Case Th ree
can
total number of ways of ac hieving t he 2 - 2 - 1 o p tion is
b. Number
= =
ways
of
achieving
t he
second
o p tion
= 15 ways. 3
-
1
-
total number of ways of getting t he 3 - 1 - 1 o p tion is 5C3 = 10 = 10 ways.
ways
number
in
w hic h
of ways 15
of
the
5
ac hieving +
toys Case
can a 10
+
be
p acked
number
of
in ways
=
3
identical
of achieving 25
bo x es Case b ways.
uestion 3 There are 2 brothers among a group of 20 persons. In how many w ays can the group be arranged around a circle so that there is exactly one person between t he two brothers? (1) (2) (3) (4) (5)
2* 18! 19! 2* 2*
1
toys out of t he 5 can be selected in 5C3 ways. As t he bo x es are identical, t he remaining two toys go into t he two identical looking bo x es in only one way.
Th erefore, T otal
of
=
17! * 18 * 18 18! 17! * 17!
Correct Answer is 2 * 18! - C hoice (4)
Exp lanation Circular Permutation 'n' objects can be arranged around a circle in (n - 1)!. If arranging these 'n' objects clockwise or counter clockwise means one a nd the same, then the number arrangements will be half that number.
i.e.,, number of arrangements = i.e.
.
Let there be exactly one person between the two brothers as stated in the question. If we consider the two brothers and the person in between between the brothers as a block, th en there will 17 others and this block of three people to be arranged around a circle. The number of ways of arranging 18 objects around a circle is in 17! ways. Now the brothers can be arranged on either side of the person who is in between the brothers in 2! ways.
The person who sits in between the two brothers could be any of the 18 in the group and can be se lected in 18 ways. Therefore, the total number of ways 18 * 17! * 2 = 2 * 18!.
Question 4 In how many ways can 15 people be seated around two rou nd tables with seating capacities of 7 and 8 people? (1) (2) (3) (4) (5)
15!/(8!) 7!*8! (15C8)*6!*7! 2*(15C7)*6!*7! 15C8 * 8!
Correct Answer is (15C8)*6!*7! - (3)
Exp lanation Circular 'n'
Permutation objects
can
be
arranged
around
a
circle
in
(n
-
1)!.
If arranging t hese 'n' objects clockwise or counter clockwise means one and t he same, t hen t he number arrangements will be t hat number. h alf
i.e.,
number
You
can
c h oose
of the
7
arrangements p eo ple
to
sit
= in
t he
. first
table
in
15
C 7 ways.
After selecting 7 p eo ple for t he table t hat can seat 7 p eo ple, t hey ca n be se ated in (7-1)! = 6!. Th e
remaining 8 p eo p le can be made to sit around t he seco nd circular table in (8-1)! = 7! Way s.
Hence,
total
number
of
ways
15
C8 *
6!
*
Permutation : Rearranging Letters of a Word Question 5 In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?
(1) (2) 3!*3!
(3)
(4) (5) 72
Correct Answer is
- C hoice 4
Exp lanation ABACUS is a 6 letter word with 3 of the letters being vowels. If the 3 vowels have to appear together as stated in the question, then there will 3 consonants and a set of 3 vowels
7!
grouped together. One group of 3 v owels and 3 consonants are e ssentially 4 elements to be rearranged. The number of possible rearrangements is 4! The group of 3 vowels contains two 'a's and one 'u'.
The 3 vowels can rearrange amongst themselves in
ways as the vowel "a" appears twice.
Hence, the total number of rearrangements in which the vowels appear together are
Question 6 There are 6 boxes numbered 1, 2,...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes c ontaining green balls are consecutively numb ered. The total number of ways in which this can be done is ____ (1) (2) (3) (4) (5)
5 21 33 60 27
Correct Answer is 21 - C hoice 2
Exp lanation If only one of the boxes has a green ball, it can be any of the 6 boxes. So, this can be achieved in 6 ways. If two of the boxes have green balls and then there are 5 arrangement possible. i.e., the two boxes can one of 1 -2 or 2-3 or 3-4 or 4-5 or 5-6. If 3 of the boxes have green balls, there will be 4 options in which the 3 boxes are in consecutive positions. i.e., 1-2-3 or 2-3-4 or 3-4-5 or 4-5-6 If 4 boxes have green balls, there will be 3 options. i.e., 1-2-3-4 or 2-3-4-5 or 3-4-5-6 If 5 boxes have green balls, then there will be 2 options. i.e., 1-2-3-4-5 or 2-3-4-5-6 If all 6 boxes ha ve green balls, then there will be just 1 options. Total number of options = 6 + 5 + 4 + 3 + 2 + 1 = 21.
Number of 4-digit numbers not greater t han 4000 The question given below appeared in CA T 2008. One could get the answer to the question in a little less than 2 minutes. What is essential is a little care to ensure that you do not make any error in oversight (commonly referred as "silly m istake"). The common error in oversight made is not reading the question correct. So, do not lose focus while reading the question. Understand it properly and then answer the question.
Question 7 How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?
(1) (2) (3) (4) (5)
499 500 375 376 501
Correct Answer is 376 - C hoice (4)
Exp lanation Th e
smallest
Th e
largest
number
number
in
in
t he
t he
series
is
series 4000,
the
is only
10 00, 4-digit
a num ber
4-digit to
sta rt
num ber. wit h
4.
Th e
left most digit (t h ousands p lace) of eac h of t he 4 digit numbers ot her t han 4000 can take one of t he values 1 or 2 or 3. Th e ne x t 3 digits ( h undreds, tens and units place) can take any of t h e 5 values 0 or 1 or 2 or 3 or 4. 3
Hence, there Including
are 3 4000,
*
5 * there
5
* 5 will
or
375 be
numbers 376
from 1000 suc h
to 3999. numbers.
Rearranging Letters of a word with constraint Question 8 In how many rearrangements of the word AMAZED, is the letter 'E' positioned in between the 2 'A's (Not necessarily flanked)? (A) 24 (B) 72 (C) 120 (D) 240 Correct Answer is 120 - C hoice (C)
Exp lanation In any rearrangement of the word, consider only the positions of the letters A, A and E. These can be as A A E, A E A or E A A. So, effectively one-third of all words will have 'E' in between the two 'A's.
The total number of rearrangements are
= 360.
One-third of 360 is 120. Answer (C) Question: A college has 10 ba sketball players. A 5-member tea m and a captain will be selected out of these 10 players. How many different selections can be made?
1. 2. 3. 4.
1260 210 10C6 * 6! 10C5 * 6
Correct Answer - 1260. Choice (1) Ex planatory Answer
A team of 6 me mbers has to be selected from the 10 players. This can be done in 10C6 or 210 ways.
Now, the captain can be selected from these 6 players in 6 ways. Therefore, total ways the selection can be made is 210*6 = 1260. Alternatively, we can select the 5 member team out of the 10 in 10C5 ways = 252 ways. The captain can be selected from amongst the remaining 5 players in 5 ways. Therefore, total ways the selection of 5 players and a captain can be made = 252*5 = 1260.
uestion:
Badri has 9 pairs of dark Blue socks and 9 pairs of Black socks. He keeps th em all in a same bag. If he picks out three socks at random what is the probability he will get a matching pair? (1) (2* 9 C2 * 9C1) / 9
9
(2) ( C2 * C1) / (3) 1
18
18
C3
C3
(4) None of these Correct Answer - (3) Solution:
If he picks any of the three so cks invariably any two of them should match. Hence the prob ability is 1. Question:
How many words of 4 consonants and 3 vowels can be made from 12 co nsonants and 4 vowels, if all the letters are different? (1)
16
C7 * 7!
12
(2) C4 * 4C3 * 7! (3) 12C3 * 4C4 (4) 12C4 * 4C3 Correct Answer - (2) Solution:
4 consonants out of 12 can be selected in 12C4 ways. 3 vowels can be selected in 4C3 ways. Therefore, total number of groups each containing 4 conso nants and 3 vowels = 12C4 * 4C3 Each group contains 7 letters, which can be arranging in 7! ways. Therefore required number of words = 124 * 4C3 * 7! Question:
If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank o f the word CHASM?
(1) 24 (2) 31 (3) 32 (4) 30 Correct Answer - (3) Solution:
The 5 letter word can be rearranged in 5! Ways = 120 without any of the letters repeating. The first 24 of these words will start with A. Then the 25th word will start will CA _ _ _. The remaining 3 letters can be rearranged in 3! Ways = 6. i.e. 6 words exist that start with CA. The next word starts with CH and then A, i.e., CHA _ _. The first of the words will be CHAMS. The next word will be CHASM. Therefore, the rank of CHASM will be 24 + 6 + 2 = 32. Question:
How many four letter distinct initials can be formed using the alphabets of English language suc h that the last of the four words is always a conson ant? (1) (26^3)*(21) (2) 26*25*24*21 (3) 25*24*23*21 (4) None of these. Correct Answer - (1) Solution:
The last of the four letter words should be a c onsonant. Therefore, there are 21 options. The first three letters can be either conso nants or vowels. So, each of them h ave 26 options. Note that the question asks you to find o ut the number of distinct initials and not initials where the letters are distinct. Hence answer = 26*26*26*21 = 26 3 * 21 Question: When four fair dice are rolled simultaneously, in how many out comes will at least one of the dice show 3? (1) 155 (2) 620
(3) 671 (4) 625 Correct Answer - (3) Solution:
When 4 dice are rolled simultaneously, there will be a total of 64 = 1296 outcomes. The number of outcomes in which none of the 4 di ce show 3 will be 54= 625 outcomes. Therefore, the number of outcomes in which at least one die will show 3 = 1296 ² 625 = 671 Question:
In how many ways can the letters of the word EDUCATION be rearranged so that the relative position of the vowels and co nsonants remain the same as in the word EDUCATION? (1) 9!/4 (2) 9!/(4!*5!) (3) 4!*5! (4) None of these Correct Answer - (3) Solution:
The word EDUCATION is a 9 letter word, with none of the letters repeating. The vowels occupy 3, 5, 7th and 8th position in the word and the remaining 5 positions are occ upied by consonants As the relative position of the vowels and conso nants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the aforementioned 4 places and the consonants can occupy 1st, 2nd, 4th, 6th and 9th positions. The 4 vowels can be arranged in the 3rd, 5th , 7th and 8th position in 4! Ways. Similarly, the 5 consonants can be arranged in 1st, 2nd, 4th , 6th and 9th position in 5! Ways. Hence, the total number of ways = 4! * 5!. Question: How many ways can 10 letters be posted in 5 post box es, if each of the post boxes can take more than 10 letters? (1) 5^10 (2) 10^5 (3) 10P5 (4) 10C5
Correct Answer - (1) Solution:
Each of the 10 letters can be pos ted in any of the 5 bo xes. So, the first letter has 5 options, so does the sec ond letter and so on and so forth for all of the 10 letters. i.e. 5*5*5*«.*5 (upto 10 times) = 5 ^ 10. A team of 8 students goes on an excursion, in two cars, of which on e can seat 5 and the other only 4. In how many ways can they travel?
(1) 9 (2) 26 (3) 126 (4) 3920 Correct Answer - (3) Solution:
There are 8 students and the m aximum capacity of th e cars together is 9. We may divide the 8 students as follows Case I: 5 students in the first car and 3 in the second Or Case II: 4 students in the first car and 4 in the second Hence, in Case I: 8 students are divided into groups of 5 and 3 in 8C3ways. Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4 ways. Therefore, the total number of ways in which 8 stud ents can travel is8C3 + 8C4 = 56 + 70 = 126. Question:
How many ways can 4 prizes be given away to 3 boys, if each boy is eligible for all the prizes? (1) 256 (2) 12 (3) 81 (4) None of these
Correct Answer - (3) Solution:
Any one prize can be given to any one of the 3 boys and hence there are 3 ways of distributing each prize. Hence, the 4 prizes can be distributed in 3 4= 81 ways.
Question:
There are 12 yes or n o questions. How many ways can these be answered? (1) 1024 (2) 2048 (3) 4096 (4) 144 Correct Answer - (3) Solution:
Each of the questions can be answered in 2 ways (yes or no) Therefore, no. of ways of answering 12 questions = 212 = 4096 ways. Question:
How many words can be formed by re-arranging the letters of the word ASCENT such that A and T occupy the first and last position respectively? (1) 5! (2) 4! (3) 6! - 2! (4) 6! / 2! Correct Answer - (2) Solution:
As A and T should occupy the first and last position, the first and last position can be filled in only one way. The remaining 4 positions can be filled in 4! Ways by th e remaining words (S,C,E,N,T). hence by rearranging the letters of th e word ASCENT we can form 1x4! = 4! words. Question:
Four dice are rolled simultaneously. What is the number of possible o utcomes in which at least one of the die shows 6? (1) 6! / 4! (2) 625
(3) 671 (4) 1296 Correct Answer - (3) Solution:
When 4 dice are rolled simultaneously, there are 64 = 1296 outcomes. The converse of what is asked in the question is that none of the dice s how '6'. That is all four dice show any o f the other 5 numbers. That is possible in 54 = 625 outcomes. Therefore, in 1296 - 625 = 671 outcomes at least one of the dice will show 6.
Question:
How many alphabets need to be there in a langu age if one were to make 1 million distinct 3 digit initials using the alphabets of the language? (1) 26 (2) 50 (3) 100 (4) 1000 Correct Answer - (3) Solution:
1 million distinct 3 digit initials are needed. Let the number of required alphabets in the langu age be ¶n·. Therefore, using ¶n· alphabets we can form n * n * n = n3 distinct 3 digit initials. Note distinct initials is different from initials where the digits are different. For instance, AAA and BBB are acc eptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different. This n3 different initials = 1 million i.e. n3 = 106 (1 million = 10 6) => n3 = (102)3 => n = 102 = 100 Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.
In how many ways can the letters of t he word MANAGEMENT be rearranged so that the two As do not appear together? (1) 10! - 2! (2) 9! - 2!
(3) 10! - 9! (4) None of these Correct Answer - (4) Solution:
The word MANAGEMENT is a 10 letter word. Normally, any 10 letter word can be rearranged in 10! ways. However, as there are certain letters of the word repeating, we need to account for thos e. In this case, the letters A, M, E and N repeat twice each. Therefore, the number of ways in w hich the letters of the word MANAGEMENT can be rearranged reduces to
.
The problem requires us to find o ut the number of o utcomes in which the two As do not appear together. The number of outcomes in which the two As appear together can be found out by considering the two As as one single letter. Therefore, there will now be only 9 letters of which three of them E, N and M repeat twice. So these 9 letters with 3 of them repeating twice can be rearranged in
ways.
Therefore, the required answer in which the two As do not appear next to each other = Total number of outcomes - the number of outcomes in which the 2 As appear toge ther
=>
ways.
Question:
There are 5 Rock so ngs, 6 Carnatic songs and 3 Indi po p songs. How many different albums can be formed using the above repertoire if the albums should contain at least 1 Rock song and 1 Carnatic song? (1) 15624 (2) 16384 (3) 6144 (4) 240 Correct Answer - (1) Solution:
There are 2n ways of choosing ¶n· objec ts. For e.g. if n = 3, then the three objects can b e chosen in the following 23 ways - 3C0 ways of choosing none of the three, 3C1 ways of choosing one out of the three,3C2 ways of choosing two out of the thre e and 3C3 ways of choosing all three.
In the given problem, there are 5 Rock songs. We can choose them in 25 ways. However, as the problem states that the case where you do not choose a Rock song does not exist (at least one rock song has to be selected), it can be do ne in 25 - 1 = 32 - 1 = 31 ways. Similarly, the 6 Carnatic songs, choosi ng at least one, can be selected in 26 - 1 = 64 - 1 = 63 ways. And the 3 Indi pop can be selected in 23 = 8 ways. Here the option of not selecting even o ne Indi Pop is allowed. Therefore, the total number of combinations = 31 * 63 * 8 = 15624 Question:
What is the value of 1*1! + 2*2! + 3!*3! + ............ n*n!, where n! means n factorial or n(n-1)(n-2)...1 (1) n(n-1)(n-1)! (2) (n+1)!/(n(n-1)) (3) (n+1)! - n! (4) (n + 1)! - 1! Correct Answer - (4) Solution:
1*1! = (2 -1)*1! = 2*1! - 1*1! = 2! - 1! 2*2! = (3 - 1)*2! = 3*2! - 2! = 3! - 2! 3*3! = (4 - 1)*3! = 4*3! - 3! = 4! - 3! .. .. .. n*n! = (n+1 - 1)*n! = (n+1)(n!) - n! = (n+1)! - n! Summing up all these terms, we ge t (n+1)! - 1! Question:
How many number of times will the digit ¶7' be written when listing the integers from 1 to 1 000? (1) 271 (2) 300 (3) 252 (4) 304 Correct Answer - (2) Solution:
7 does not occ ur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0< x, y, z < 9. 1. The numbers in which 7 occurs o nly once. e.g 7, 17, 78, 217, 743 etc This means that 7 is one of the digits and the rem aining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)
You have 1*9*9 = 81 such numb ers. However, 7 could appear as the first or the second o r the third digit. Therefore, there will be 3*81 = 243 n umbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once. In each of these numbers, 7 is written once. Therefore, 243 times. 2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77 In these numbers, one of the digits is not 7 and it c an be any of the 9 digits ( 0 to 9 with the exception of 7). There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers. In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times. 3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it. Therefore, the total number of times the digit 7 is written between 1 and 999 is 243 + 54 + 3 = 300
FACTORIAL 'n'
The continuous product of the first 'n' natural numbers is called factorial n and is deonoted by n! i.e, n!=1ò2ò 3x⼦..x(n-1)xn. PERMUTATION
An arrangementthat can be formed by taking some or all of a finite set of things (or objects) is called a Permutation.Order of the things is very important in case of permutation.A permutation is said to be a Linear Permutation if the objects are arranged in a line. A linear permutation is simply called as a permutation.A permutation is said to be a Circular Permutation if the objects are arranged in the form of a circle.The number of (linear) permutations that can be formed by taking r things at a time from a set of n distinct things denoted by
is
.%
NUMBER OF PERMUTATIONS UND ER CERTAIN CONDITIONS
1. Number of permutations of n different things, taken r at a time, when a particular thng is to be always included in each arrangement , is
.
2. Number of permutations of n different things, taken r at a time, when a particular thing is never taken in each arrangement is . 3. Number of permutations of n different things, taken all at a time, when m specified things always come together is
.
4. Number of permutations of n different things, taken all at a time, when m specified never come together is
.
5. The number of permutations of n dissimilar things taken r at a time when k(< r) particular things always occur is
.
6. The number of permutations of n dissimilar things taken r at a time when k particular things never occur is . 7. The number of permutations of n dissimilar things taken r at a time when repetition of things is allowed any number of times is
8.
The number of permutations of n different things, taken not more than r at a time, when each thing may occur
any number of times is
.
9. The number of permutations of n different things taken not more than r at a time . *PERMUTATIONS OF SIMILAR THINGS*The numb er of per mutations of n t hings tak en all tat a time when p of t hem are all
alik e and the rest are all diff erent is
.If p things are alik e of one type, q things are alik e of other type, r things are alik e of
another type, then the number of per mutations with p+q+r things is
.
CIRCULAR PERMUTATIONS
}1. The number of circular permutations of n dissimilar things taken r at a time is
.
2. The number of circular permutations of n dissimilar things taken all at a time is
.
3. The number of circular permutations of n things taken r at a time in one direction is
.
4. The number of circular permutations of n dissimilar things in clock-wise direction = Number of permutations in anticlock-wise direction =
.
COMBINATION
A selection that can be formed by taking some or all of a finite set of things( or objects) is called a Combination The number of combinations of n dissimilar things taken r at a time is denoted by
.
1. 2. 3. 4. 5. The number of combinations of n things taken r at a time in which a)s particular things will always occur is b)s particular things will never occur is
. .
c)s particular things always occurs and p particular things never occur is
.
DISTRIBUTION OF THINGS INTO GROU PS
1.Number of ways in which (m+n) items can be divided into two unequal groups containing m and n items is
.
2.The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is 3.The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important is
.
4.The number of ways in which (m+n+p) things can be divided into three different groups of m,n, an p things respectively is 5.The required number of ways of dividing 3n things into three groups of n each =
.When the order of
groups has importance then the required number of ways= DIVISION OF ID ENTICAL OBJ ECTS INTO GROU PS
The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items
is
}The number of non-negative integral solutions of the equation . The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is The number of positive integral solutions of the equation . The number of ways of choosing r objects from p objects of one kind, q objects of second kind, and so on is the coefficient of in the expansion
he number of ways of choosing r objects from p objects of one kind, q objects of second kind, and so on, such that one object of each kind may be included is the coefficient of is the coefficient of in the expansion . %{font-family:ver dana}+*TOTAL NUMBER OF COMBINATIONS*+% %{font-family:ver dana}1.The total numb er of combinations of when things are alik e of one kind , things are alik e of second kind⼦. is
things tak en any number at a time things are alik e of kind,
.%
%{font-family:ver dana}2.The total numb er of combinations of when things are alik e of on e kind, things are alik e of second kind⼦.
things tak en one or more at a time things are alik e of kind, is%
. SUM OF THE NUMBERS
Sum of the numbers formed by taking all the given n digits (excluding 0) is Sum of the numbers formed by taking all the given n digits (including 0) is
Sum of all the r-digit numbers formed by taking the given n digits(excluding 0) is
%
%{font-family:verdana}Sum of all the r-digit numbers formed by taking the given n digits(including 0) is
DE-ARRANG EMENT:
The number of ways in which exactly r letters can be placed in wrongly addressed envelopes when n letters are placed in n addressed envelopes is
.
The number of ways in which n different letters can be placed in their n addressed envelopes so that al the letters are in the wrong envelopes is
.
IMPORTANT RESULTS TO R EMEBER
In a plane if there are n points of which no three are collinear, then 1. The number of straight lines that can be formed by joining them is 2. The number of triangles that can be formed by joining them is
. .
3. The number of polygons with k sides that can be formed by joining them is
.
In a plane if there are n points out of which m points are collinear, then 1. The number of straight lines that can be formed by joining them is 2. The number of triangles that can be formed by joining them is
. .
3. The number of polygons with k sides that can be formed by joining them is
.
Number of rectangles of any size in a square of n x n is In a rectangle of p x q (p < q) number of rectangles of any size is In a rectangle of p x q (p < q) number of squares of any size is n straight lines are drawn in the plane such that no two lines are parallel and no three lines three lines are concurrent. Then the number of parts into which these lines divide the plane is equal to
.
In how many ways can the letters of the word ³PROBLEM´ be rearranged to make 7 letter words such that none of the letters repeat? (1)
7!
(2)
7C7
(3)
7
(4)
49
(5)
None of these
7
Correct Answer Choice (1)
Solution:
There are seven positions to be filled. The first position can be filled using any of the 7 letters contained in PROBLEM. The second position can be filled by the remaining 6 letters as the letters should not repeat. The third position can be filled by the remaining 5 letters only and so on.
Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! Ways.
en coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head? (A)
210
(B)
29
(C)
3*28
(D)
3*29
(E)
None of these
Correct Answer Choice (B)
Solution:
When a coin is tossed once, there are two outcomes. It can turn up a head or a tail. When 10 coins are tossed simultaneously, the total number of outcomes = 2
10
Out of these, if the third coin has to turn up a head, then the number of possibilities for the third coin is only 1 as the outcome is fixed as head. 9 Therefore, the remaining 9 coins can turn up either a head or a tail = 2 In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes? (A)
5
(B)
5
C
3
P
3
3
(C)
5
(D)
35
(E)
25
5
Correct choice (D) Correct Answer (3 )
Solution:
Explanatory Answer The first letter can be posted in any of the 3 post boxes. Therefore, it has 3 choices. Similarly, the second, the third, the fourth and the fifth letter can each be posted in any of the 3 post boxes. Therefore, the total number of ways the 5 letters can be posted in 3 boxes is 3*3*3*3*3 =3
5
A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target ? (A)
1
(B)
(C)
(D)
(E)
Correct Choice (D) Answer (
)
Explanatory Answer The man will hit the target even if he hits it once or twice or thrice or all four times in the four shots that he takes. So, the only case where the man will not hit the target is when he fails to hit the target even in one of the four shots that he takes.
The probability that he will not hit the target in one shot = 1 -
=
Therefore, the probability that he will not hit the target in all the four shots =
Hence, the probability that he will hit the target at least in one of the four shots = 1 -
=
.
here are 6 boxes numbered 1, 2,....6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is (A)
5
(B)
21
(C)
33
(D)
60
(E)
6
Correct Choice (B) Answer (21)
Explanatory Answer If only one of the boxes has a green ball, it can be any of the 6 boxes. So, this can be achieved in 6 ways. If two of the boxes have green balls and then there are 5 consecutive sets of 2 boxes. 12, 23, 34, 45, 56. Similarly, if 3 of the boxes have green balls, there will be 4 options. If 4 boxes have green balls, there will be 3 options. If 5 boxes have green balls, then there will be 2 options. If all 6 boxes have green balls, then there will be just 1 options. Total number of options = 6 + 5 + 4 + 3 + 2 + 1 = 21.
Question What is the probability that the position in which the consonants appear remain unchanged when the letters of the word Math are re-arranged? A.
1/4
B.
1/6
C.
1/3
D.
1/24
E.
1/12
Correct Choice (A) Answer (1/4)
Explanatory Answer
The total number of ways in which the word Math can be re-arranged = 4! = 4*3*2*1 = 24 ways. Now, if the positions in which the consonants appear do not change, the first, third and the fourth positions are reserved for conso nants and the vowel A remains at the second position. The consonants M, T and H can be re-arranged in the first, third and fourth positions in 3! = 6 ways without the positions in which the positions in which the consonants appear changing.
Therefore, the required probability = How many different four letter words can be formed (the words need not be meaningful) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?
A.
59
B. C.
56
D.
23
E. Correct Choice (A) Answer (59)
Explanatory Answer
The first letter is E and the l ast one is R. Therefore, one has to find two more letters from the remaining 11 letters. Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters. The second and third positions can either have two different letters or have both the letters to be the same. Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8*7 = 56 ways. Case 2: When the two letters are same. There are 3 options ± the three can be either Ns or Es or As. Therefore, 3 ways. Total number of posssibilities = 56 + 3 = 59
Permutations
Loosely speaking, permutations are used to display the total number of outcomes given a certain fact. In permutations, order matters. Permutations can be calculated using the formula
Combinations Combinations are very similar to permutations. The only difference is that in combinations, order does not matter. The formula to calculate the number of possible combinations is:
Understanding the specific differences between permut ations and combinations is not important. Its only important to know which equation to apply depending on the problem statements. E xampl e
A thief is trying to crack an atm machine. The keypad consists of 10 numbers (0-9). If no numbers can be repeated and the atm password is 4 characters long, how many unique combinations exist? First determine whether this is a permutation or combination question. Since changing the order of numbers gives us a new combin ation, order matters. This is a permutation question. n = 10, because the thief is choosing from a total of 10 possible numbers. k = 4, because each atm password is 4 characters long. Apply the formula :
There ar e 5040 possi bl e com binations E xampl e
A manag er is s el ecting 3 p eopl e out of a pool of 7 volunt eers to form a n ew saf ety committ ee. How many diff er ent committ ees can be form ed? Det ermin e wh eth er this is a p ermutation or com bination pro bl em . Sinc e changing th e ord er of th e m em bers on a t e am would still gi ve us th e sam e t eam, ord er do e s not matt er . T his is a com bination pro bl em . n = 7, becaus e th er e ar e a total of 7 volunt eers k = 3, becaus e w e ar e trying to form a committ ee of 3 m em bers
T he r e ar e 35 diff er ent committ ees that can be form ed
Advanced Combination /
Permutations
The most difficult combination and permutation problems involve constraints. In these advanced difficulty problems, a best practice approach is to list the winning scenarios. There are two rules to remember: 1. 2.
Y ou Y ou
multiply when you take a sequence of actions to get the desired outcome. add when you break the outcome into different cases. Y ou add the cases to get the total.
E xampl e
From a group of 7 m en and 5 wom en, how many diff er ent committ ees consisting of 4 m en and 2 wom en can be form ed? Det ermin e wh eth er this is a p ermutation or com bination pro blem . Sinc e changing th e ord er of th e m em bers on a committ ee wou ld sti ll gi ve us th e sam e committ ee, ord er do es not matt er . This is a com bination pro blem . This pro blem fo llows a s equ enc e of actions : first choos e th e m en, and th en choos e th e wom en . App ly this s equ enc e into th e com binations formu l a :
There are 350 committees that can be formed. E xampl e
From a g roup of 7 men and 5 w omen, how many different committees consisting of 4 men and 2 w omen can be formed if Jenn y and John ref u se to be on the committee tog ether? Determine w hether this is a p ermu tation or combination p roblem. Since chang ing the order of the members on a committee w ou ld still g ive u s the same committee, order does not matter. This is a combination p roblem. Deconstru ct the p roblem into different cases and s u m tog ether the p ossible combinations of all the different cases to find the total n u mber of committees p ossible. Case 1: Both Jenny and John are not on the committee. Sequ ence of actions: choose 4 men from the remainin g 6 men; choose 2 w omen from the rem aining 4 w omen
Case 2: Jenny is on the committee, and John is not.
Case 3: John is on the committee, and Jenny is not.
Total: 90 + 60 + 120 = 270 1. A committee of three must be formed from 5 women and 5 men. What is the probability that the committee will be exclusive to one gender? (A) 1/60 (B) 1/120
(C) 1/8 (D) 1/6 (E) 1/3
2. A three-letter code is formed using the letters A-L, such that no letter is used more than once. What is the probability that the code will have a string of three consecutive letters (e.g. A-B-C, F-E-D)? (A) 1/55 (B) 1/66 (C) 2/17 (D) 1/110 (E) 2/55
3. A homework assignment calls for students to write 5 sentences using a total of 10 vocabulary words. If each sentence must use two words and no words can be us ed more than once, then how many different ways can a student select the words? (A) 10!/5! (B) 10!/32 (C) 5! x 5! (D) 2! x 5! (E) 10!
4. Team S is to comprise of n debaters chosen from x people? Team R is comprised of n + 1 debaters chosen from x+1 people. Column A Number of unique team S
Column B Number of unique Team R
5. A lunar mission is made up of x astronauts and is formed from a total of 12 astronauts. A day before the launch the commander of the program decides to add p astronauts to the mission. If the total number of possible lunar missions remain unchanged after the commander¶s decision, then which of the following cannot be the value of p? (A) x (B) x + 3 (C) 3 (D) 6 (E) 8