# Permutation and Combination Quiz 10

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# Permutation and Combination Quiz 10

### Introduction

Permutation and Combination is one of important topic in Quantitative Aptitude Section. In Permutation and Combination Quiz 10 article candidates can find questions with an answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Permutation and Combination Quiz 10 questions are very useful for different exams such as IBPS PO, Clerk, SSC CGL, SBI PO, NIACL Assistant, NICL AO, IBPS SO, RRB, Railways, Civil Services etc.

### Q1

There are 12 boys and 15 girls. How many different dancing groups can be formed with 2 boys and 3 girls?
A. 30030 B. 35060 C. 35565 D. 31084

A
Selecting 2 boys from a group of 12 and correspondingly selecting 3 girls from group of 15 girls
$^{12}{C}_{2} \times ^{15}{C}_{3} = 30030$
i.e, $^{n}{C}_{r} = \frac{n!}{r! \times (n-r)!}$

### Q2

A four letter code has to be formed using the alphabets form the set {a, b, c, d} such that the codes formed have odd number of a's and other alphabets cannot be repeated. How many different codes can be formed
A. 24 B. 120 C. 36 D. 30

C
Since the code has to be four lettered and need to have odd number of a’s, the code can have either 1 a or 3 a’s.
If the code has one a, then the remaining letters of the code are b, c and d.
∴ Number of different possible codes = 4! = 24
If the code has three a’s then remaining letter can be chosen in 3c1 = 3 ways
∴ Number of ways of forming the code = 3 × (4! ÷ 3!) = 12 (∵ 3 letters are repeated)
∴ Total number of ways of forming the code = 24 + 12 = 36 ways.

### Q3

23 people are there, they are shaking hands together. How many handshakes possible, if they are in pair of cyclic sequence?
A. 250 B. 252 C. 242 D. 253

D
Every person shakes hand with every other person = 22 hand shakes
⇒ Total handshakes = 22 × 23
But A handshaking with B is same as B handshaking A
⇒ Total handshakes = $\frac{(22 × 23)}{2}$ = 253

### Q4

How many alphabets need to be there in a language if one were to make 1 million 3 letter initials using the alphabets of the language?
A. 140 B. 1000 C. 10 D. 100

D
Suppose we have A number of alphabets in a language.
⇒ Number of ways to make a 3 letter initial = A × A × A = ${A}^{3}$
Given, number of ways to make 3 letter initial = 1 million = ${10}^{6}$
⇒ ${A}^{3}$ = ${10}^{6}$
⇒ A = ${10}^{2}$ = 100
∴ The language needs 100 alphabets

### Q5

Six friends go to pizza corner and there are 2 types of pizzas and six different flavors for each type. They have to select 2 flavors of same type from 6 flavors. In how many ways they can select pizza?
A. 25 ways B. 20 ways C. 30 ways D. 45 ways

C
Number of ways 2 flavors can be selected out of 6 flavors = $^{6}{C}_{2}$ = 15
Number of ways pizza can be selected = 2 [there are 2 types of pizzas]
∴ Number of ways to select a pizza = 2 × 15 = 30 ways

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