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Chapter 5 – Permutations and
Combinations

CA NISHANT KUMAR 1

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Question 1
Find x if 1/9! + 1/10! = x/11!
(a) 119

(b) 120

(c) 121

(d) None

CA NISHANT KUMAR 8

Question 1

CA NISHANT KUMAR 9

Question 2

( n + 1)! − n! = ?
(a) n
...
n

(d) None

CA NISHANT KUMAR 10

Question 3
The number of ways the letters of the word ‘COMPUTER’ can be rearranged is:
(a) 40,320

(b) 40,319

(c) 40,318

(d) None

CA NISHANT KUMAR 11

Question 4
If 18 Cr = 18Cr + 2 , the value of r C5 is:
(a) 55

(b) 50

(c) 56

(d) None

CA NISHANT KUMAR 12

Question 5
In n Pr , the restriction is:
(a) n > r

(b) n ≥ r

(c) n ≤ r

(d) None

CA NISHANT KUMAR 13

Question 6
In n Pr = n ( n − 1)( n − 2 )
...

(a) 20

(b) 21

(c) 22

(d) None

CA NISHANT KUMAR 15

Question 8
In a group of boys the number of arrangement of 4 boys is 12 times the number of
arrangements of 2 boys
...
Find the number of ways of forming
such a committee
...
In how many different ways can the prizes be awarded?
(a) 1,716

(b) 1,720

(c) 1,270

(d) None

CA NISHANT KUMAR 19

Question 12
In how many different ways can 3 students be associated with 4 chartered accountants,
assuming that each chartered accountant can take at most one student?
(a) 24

(b) 25

(c) 30

(d) None

CA NISHANT KUMAR 20

Question 13
When Dr
...
If he
can see only one patient at a time, find the number of ways, he can schedule his patients
(a) if they all want their turn, and (b) if 3 leave in disgust before Dr
...

(a) 12 P12 , 12 P9

(b) 12!, 12 P9

(c) Both

(d) None

CA NISHANT KUMAR 21

Question 14
Mr
...
Y enter into a railway compartment having six vacant seats
...
The number of arrangements is:
(a) 9 8

(b) 10

(c) 8 9

(d) None

CA NISHANT KUMAR 24

Question 17
n articles are arranged in such a way that 2 particular articles never come together
...
The number of such is:
(a) 72

(b) 27

(c) 70

(d) None

CA NISHANT KUMAR 26

Question 19
The number of ways the letters of the word ‘TRIANGLE’ to be arranged so that the word
‘angle’ will be always present is:
(a) 20

(b) 60

(c) 24

(d) 32

CA NISHANT KUMAR 27

Question 20
If the letters word ‘DAUGHTER’ are to be arranged so that vowels occupy the odd
places, then number of different words are:
(a) 2,880

(b) 676

(c) 625

(d) 576

CA NISHANT KUMAR 28

Question 21
How many arrangements can be made out of the letters of the word ‘DRAUGHT’, the
vowels never beings separated?
(a) 1,440

(b) 676

(c) 625

(d) 576

CA NISHANT KUMAR 29

Question 22
There are 6 books on Economics, 3 on Mathematics and 2 on Accountancy
...
In
how many ways can they be seated if (i) all the sisters sit together, (ii) no two sisters sit
together?
(a) 720; 1,440

(b) 1,440; 720

(c) 840; 720

(d) None

CA NISHANT KUMAR 31

Question 23

CA NISHANT KUMAR 32

Question 24
Six boys and five girls are to be seated for a photograph in a row such that no two girls
sit together and no two boys sit together
...

(a) 72,000

(b) 14,440

(c) 86,400

(d) None

CA NISHANT KUMAR 33

Question 25
The number of arrangements of 10 different things taken 4 at a time in which one
particular thing always occurs is:
(a) 2015

(b) 2016

(c) 2014

(d) None

CA NISHANT KUMAR 34

Question 26
The number of permutations of 10 different things taken 4 at a time in which one
particular thing never occurs is:
(a) 3,020

(b) 3,025

(c) 3,024

(d) None

CA NISHANT KUMAR 35

Question 27
The number of numbers lying between 10 and 1000 can be formed with the digits 2, 3,
4, 0, 8, 9 is:
(a) 124

(b) 120

(c) 125

(d) None

CA NISHANT KUMAR 36

Question 28
The total number of 9 digit numbers of different digits is:
(a) 10 9

(b) 8 9

(c) 9 9

(d) None

CA NISHANT KUMAR 37

Question 29
There are 5 speakers A, B, C, D and E
...
The number of ways in which a
person can go from Calcutta to Delhi and return by a different train is:
(a) 99

(b) 90

(c) 80

(d) None

CA NISHANT KUMAR 39

Question 31
Number ways of painting of a face of a cube by 6 colours is:
(a) 36

(b) 6

(c) 24

(d) 20

CA NISHANT KUMAR 40

Question 32
How many six-digit telephone numbers can be formed by using 10 distinct digits?
(a) 108

(b) 610

(c) 10 C9

(d) 10 P6

CA NISHANT KUMAR 41

Question 33
The number of ways in which 8 sweats of different sizes can be distributed among 8
persons of different ages so that the largest sweat always goes to be younger assuming
that each one of then gets a sweat is:
(a) 8

(b) 5040

(c) 5039

(d) None

CA NISHANT KUMAR 42

Question 34
The number of arrangements in which the letters of the word ‘MONDAY’ be arranged
so that the words thus formed begin with M and do not end with N is:
(a) 720

(b) 120

(c) 96

(d) None

CA NISHANT KUMAR 43

Question 35
How many numbers of seven-digit numbers which can be formed from the digits 3, 4, 5,
6, 7, 8, 9, no digit being repeated are not divisible by 5?
(a) 4320

(b) 4690

(c) 3900

(d) 3890

CA NISHANT KUMAR 44

Question 35

CA NISHANT KUMAR 45

Question 36
Eight guests have to be seated 4 on each side of a long rectangular table
...
The number of ways
in which the sitting arrangements can be made is:
(a) 1732

(b) 1728

(c) 1730

(d) 1278

CA NISHANT KUMAR 46

Question 36

CA NISHANT KUMAR 47

Question 37
The number of even numbers greater than 300 can be formed with the digits 1, 2, 3, 4, 5
without repetition is:
(a) 110

(b) 112

(c) 111

(d) None

CA NISHANT KUMAR 48

Question 37

CA NISHANT KUMAR 49

Question 37

CA NISHANT KUMAR 50

Question 38
The letters of the words ‘CALCUTTA’ and ‘AMERICA’ are arranged in all possible
ways
...
The number of guests in the party is:
(a) 11

(b) 12

(c) 13

(d) 14

CA NISHANT KUMAR 53

Question 41
There are 12 points in a plane of which 5 are collinear
...
In how many ways can he invite 7 guests
such that 5 of them are relatives?
(a) 420

(b) 446

(c) 336

(d) None

CA NISHANT KUMAR 56

Question 44
The number of parallelograms that can be formed from a set of four parallel lines
intersecting another set of three parallel lines is:
(a) 6

(b) 18

(c) 12

(d) 9

CA NISHANT KUMAR 57

Question 45
The Supreme Court has given a 6 to 3 decision upholding a lower court; the number of
ways it can give a majority decision reversing the lower court is:
(a) 256

(b) 276

(c) 245

(d) 226

CA NISHANT KUMAR 58

Question 46
Five bulbs of which three are defective are to be tried in two bulb points in a dark room
...
The number of committees
such that each committee includes at least one lady is:
(a) 400

(b) 440

(c) 441

(d) None

CA NISHANT KUMAR 61

Question 49
There are 7 men and 3 ladies
...
In how many ways can this be done if in the committee, there
must be at least one member from each group and at least 3 Chartered Accountants?
(a) 3,570

(b) 3,750

(c) 7,350

(d) None

CA NISHANT KUMAR 64

Question 50

CA NISHANT KUMAR 65

Question 51
A committee of 3 ladies and 4 gents is to be formed out of 8 ladies and 7 gents
...
X
refuses to serve in a committee in which Mr
...
The number of such
committees is:
(a) 1530

(b) 1500

(c) 1520

(d) 1540

CA NISHANT KUMAR 66

Question 51

CA NISHANT KUMAR 67

Question 52
A boy has 3 library tickets and 8 books of his interest in the library
...
In
how many ways can he choose the three books to be borrowed?
(a) 41

(b) 51

(c) 61

(d) 71

CA NISHANT KUMAR 68

Question 52

CA NISHANT KUMAR 69

Question 53
The ways of selecting 4 letters from the word ‘EXAMINATION’ is
(a) 136

(b) 130

(c) 125

(d) None

CA NISHANT KUMAR 70

Question 53

CA NISHANT KUMAR 71

Question 53

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Question 54
The number of 4-digit numbers formed with the digits 1, 1, 2, 2, 3, 4 is:
(a) 100

(b) 101

(c) 201

(d) None

CA NISHANT KUMAR 73

Question 54

CA NISHANT KUMAR 74

Question 54

CA NISHANT KUMAR 75

Question 55
In how many ways can 8 persons be seated at a round table? In how many cases will 2
particular persons sit together?
(a) 5,040; 1,440

(b) 5,040; 720

(c) 5,040; 120

(d) None

CA NISHANT KUMAR 76

Question 56
The number of ways in which 7 boys sit in a round table so that two particular boys may
sit together is:
(a) 240

(b) 200

(c) 120

(d) None

CA NISHANT KUMAR 77

Question 57
3 ladies and 3 gents can be seated at a round table so that any two and only two of the
ladies sit together
...
Find the number of ways in which they can be seated
...
8!
(a)

7!
8!
15
(d) 2
...
The number of different
forecasts containing exactly 6 correct results is:
(a) 316

(b) 214

(c) 112

(d) None

CA NISHANT KUMAR 84

Question 62
Eight chairs are numbered from 1 to 8
...
First, the women choose the chairs from the chairs numbered
1 to 4 and then men select the chairs from the remaining
...

The maximum number of trials that are needed to assigns the keys to the corresponding
locks is:
(a)

( n −1)

C2

(b)

( n +1)

n

C2

(c)

 ( k − 1)
k =2

n

(d)

k
k =2

CA NISHANT KUMAR 86

Question 63

CA NISHANT KUMAR 87

Some Standard Results
1
( n − 1)!
...
Number of ways of selecting some or all items from a set of n items –
a
...


1
...
When there are 3 choices for each item: ( 3n − 1)
...
n
; n
Cr +1 n − r
Cr
n − r +1
3
...
If n Cx = nC y , and x  y , then x + y = n
...
If n Px = n Py , and x  y , then x + y = 2n − 1
...
The number of diagonals in a polygon of n sides is

1
n ( n − 3)
...
Division of Items in Groups –
a
...
Equal items in every group – The number of ways to divide n students
n!
into k groups of h students each is given by

...
Unequal items in every group – The number of ways to divide n items into
3 groups → one containing a items, the second containing b items, and
n!
the third containing c items, such that a + b + c = n , is given by

...
Division of Identical Items in Groups – The number of ways to divide n
n!
identical objects into k groups of h items each is given by

...
Number of Factors of a number – Factors of a number N refers to all the numbers
which divide N completely
...
qb
...

Step 2 – Use the formula: Number of factors of N = ( a + 1)( b + 1)( c + 1)
...


The maximum number of points of intersection of n circles will be n ( n − 1)
...
At least one question from each section is to be attempted
...

Therefore, 26 − 1
...
Therefore, 24 − 1
...


CA NISHANT KUMAR 93

Question 66
There are 12 questions to be answered in Yes or No
...
e
...

Therefore, all the 12 questions can be answered in 212 = 4096 ways
...
In how many different
ways can be failed?
(a) 14

(b) 16

(c) 15

(d) None

Solution
(c)
The candidate would be failed if he fails in one or more papers
...
of ways of selecting one or more items from n items is given by 2n − 1
...
of ways he can be failed = 24 − 1 = 16 − 1 = 15
CA NISHANT KUMAR 95

Question 68
In an election the number of candidates is one more than the number of members to be
elected
...

(a) 8

(b) 10

(c) 7

(d) None

Solution
(a)
In an election the number of candidates is one more than the number of members to be
elected
...
In other words, if, suppose the total number of candidates is n, then the
number of candidates to be voted for are n – 1
...

Now, number of ways of selecting one or more items from a set of n items = 2n − 1
...

Therefore, we need to subtract that one extra candidate as well
...

 2n − 2 = 254
 2n = 254 + 2

CA NISHANT KUMAR 97

 2n = 256
 2n = 28

n =8

CA NISHANT KUMAR 98

Question 69
n

C1 + nC2 + nC3 + nC4 +
...
In how many ways
can an examinee answer one or more questions?
(a) 720

(b) 728

(c) 729

(d) None

Solution
(b)
There are three choices for each question – either to attempt alternative 1, or, to attempt
alternative 2, or, to not attempt the question
...


CA NISHANT KUMAR 100

Question 71
Find x if 12 C5 + 2 12C4 + 12C3 = 14Cx
(a) 5

(b) 6

(c) 7

(d) 8

Solution
(a)
12

C5 + 2 12C4 + 12C3

12

C5 + 12C4 + 12C4 + 12C3

(

12

C5 + 12C4 ) + ( 12C4 + 12C3 )

CA NISHANT KUMAR 101

(

12 +1

13

C5 + 13C4

13+1

14

C5 ) + ( 12+1 C4 )

C5

C5

CA NISHANT KUMAR 102

Question 72
If n Cr −1 = 56 , n Cr = 28 , and n Cr +1 = 8 , find the value of r
...
(1)
n
Cr +1 n − r
n−r 8
n−r 2
n

n

Now,

Cr −1
r
56
r
r
=

=

= 2  2n − 3r = −2 …Eq
...


CA NISHANT KUMAR 104

Question 73
If n C10 = nC14 , then what is the value of
(a) 24

(b) 25

25

Cn ?
(c) 1

(d) None

Solution
(b)
We know that if n Cx = nC y , and x  y , then x + y = n
...

 25C24 = 25

CA NISHANT KUMAR 105

Question 74
If n Pr = n Pr +1 , and n Cr = nCr −1 , then find the value of n
...
Therefore, r + r + 1 = 2n − 1
...

2

CA NISHANT KUMAR 106

Also, we know that if n Cx = nC y , and x  y , then x + y = n
...

 2r − 1 = n  2r = n + 1  r =

Therefore, n − 1 =

n +1

...

2

CA NISHANT KUMAR 107

Question 75
The number of diagonals in a decagon is:
(a) 30

(b) 35

(c) 45

(d) None

Solution
(b)
The number of diagonals in a polygon of n sides is

1
n(n − 3)
...

2

1
n ( n − 3) = 44
2
CA NISHANT KUMAR 109

n ( n − 3) = 44  2 = 88
n ( n − 3) = 88

Try the options
...

k !( h!)
We have, n = 12; k = 3; h = 4
...


CA NISHANT KUMAR 112

Question 78
The number of ways in which 9 things can be divided into twice groups containing 2, 3,
and 4 things respectively is:
(a) 1250

(b) 1260

(c) 1200

(d) None

Solution
(b)
The number of ways to divide n items into 3 groups → one containing a items, the second
containing b items, and the third containing c items, such that a + b + c = n , is given by
n!

...

( h!)
We have, n = 15; k = 3; h = 5
...


CA NISHANT KUMAR 116

Question 80
The number of different factors of the number 75,600 has is:
(a) 120

(b) 121

(c) 119

(d) None

Solution
(c)

75,600 = 24
...
52
...
Since one
of the factors is the number itself, the different factors would be determined by
subtracting 1 from the total number of factors
...
of different factors of the number 75,600 = 120 – 1 = 119
...

(a) 90

(b) 100

(c) 110

(d) 120

Solution
The maximum number of points of intersection of n circles will be n ( n − 1)
...


CA NISHANT KUMAR 119



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