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Chapter 5 – Permutations and
Combinations

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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 n
handshakes is 66
...
The number of triangles is:
(a) 200

(b) 211

(c) 210

(d) None

CA NISHANT KUMAR 54

Question 42
The total number of ways in which six ‘+’ and four ‘–’ signs can be arranged in a line
such that no two ‘–’ signs occur together is:
(a) 7 / 3

(b) 6  7 / 3

(c) 35

(d) None

CA NISHANT KUMAR 55

Question 43
A person has 12 friends of whom 8 are relatives
...

Number of trials the room shall be lighted is:
(a) 6

(b) 8

(c) 5

(d) 7

CA NISHANT KUMAR 59

Question 47
The number of different words that can be formed with 12 consonants and 5 vowels by
taking 4 consonants and 3 vowels in each word is:
(a) 12 C4  5C3

(b) 17 C7

(c) 4950  7!

(d) None

CA NISHANT KUMAR 60

Question 48
Out of 7 gents and 4 ladies a committee of 5 is to be formed
...
Find the number of ways in which a committee of 6 can be
formed of them if the committee is to include atleast two ladies?
(a) 420

(b) 140

(c) 168

(d) None

CA NISHANT KUMAR 62

Question 49

CA NISHANT KUMAR 63

Question 50
A committee of 7 members is to be chosen from 6 Chartered Accountants, 4 Economists
and 5 Cost Accountants
...
Mrs
...
Y is a member
...
Of these 8, he does
not want to borrow Mathematics Part II unless Mathematics Part I is also borrowed
...
The number of ways is:
(a) 70

(b) 27

(c) 72

(d) None

CA NISHANT KUMAR 78

Question 58
5 persons are sitting in a round table in such way that Tallest Person
(a) 6

(b) 7

(c) 8

(d) 11

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

1
n ( n − 3)
...

Option (d) → 11

CA NISHANT KUMAR 110

Question 77
In how many number of ways can 12 students be equally divided into three groups?
(a) 5775

(b) 7575

(c) 7755

(d) None

Solution
(a)
The number of ways to divide n students into k groups of h students each is given by
n!
k
...

CA NISHANT KUMAR 111

Therefore,

n!
k !( h!)

k

=

12!
3!( 4!)

3

= 5,775
...

a !b!c!
CA NISHANT KUMAR 113

Here, n = 9; a = 2; b = 3; c = 4

n!
9!
=
= 1,260
a !b!c! 2!  3!  4!

CA NISHANT KUMAR 114

Question 79
In how many number of ways can 15 mangoes be equally divided among 3 students?
(a) 15 / ( 5 )

4

(b) 15 / ( 5 )

3

(c) 15 / ( 5 )

2

(d) None

Solution
(b)
The number of ways to divide n identical objects into k groups of h items each is given
n!
by
k
...

CA NISHANT KUMAR 115

Therefore,

n!

( h!)

k

=

15!

( 5!)

3

= 7,56,756
...
33
...
7
Therefore, total number of factors of 75,600 = ( 4 + 1)( 3 + 1)( 2 + 1)(1 + 1) = 120

CA NISHANT KUMAR 117

However, the question has asked us the different factors of the number 75,600
...

Therefore, no
...


CA NISHANT KUMAR 118

Question 81
The maximum number of points of intersection of 10 circles will be ________
...

Therefore, 10 (10 − 1) = 10  9 = 90

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