# Permutation and Combination – Quiz 1

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# Permutation and Combination – Quiz 1

### Introduction

Permutation and Combination is one of important topic in Quantitative Aptitude Section. In Permutation and Combination – Quiz 1 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 1 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

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
A. 564 B. 645 C. 735 D. 756

D
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
i.e, Required number of ways = $(^{7}{C}_{3} \times ^{6}{C}_{2})$ + $(^{7}{C}_{4} \times ^{6}{C}_{1})$ + $(^{7}{C}_{5})$
= $\frac {7 \times 6 \times 5}{6 \times 5} \times \frac {6 \times 5 }{2 \times 1}$ + $(^{7}{C}_{3} \times ^{6}{C}_{1})$ + $(^{7}{C}_{2})$
= 525 + $(\frac {7 \times 6 \times 5}{6 \times 5} \times 6) + (\frac {7 \times 6}{2 \times 1})$
= (525 + 210 + 21)
= 756.

### Q2

In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
A. 360 B. 480 C. 720 D. 504

C
The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.

### Q3

In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?
A. 1440 B. 2880 C. 50400 D. 810

C
In the word 'CORPORATION', we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters = $\frac {7!}{2!}$ = 2520.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in $\frac {5!}{3!}$ = 20 ways.
i.e, Required number of ways = (2520 x 20) = 50400

### Q4

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
A. 210 B. 1050 C. 25200 D. 21400

C
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
= $(^{7}{C}_{3} \times ^{4}{C}_{2})$
= $(\frac {7 \times 6 \times 5}{3 \times 2 \times 1} \times 6) + (\frac {4 \times 3}{2 \times 1})$
= 210.
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging
5 letters among themselves = 5!
= 5 x 4 x 3 x 2 x 1
= 120.
Required number of ways = (210 x 120) = 25200.

### Q5

In how many ways can the letters of the word 'LEADER' be arranged?
A. 72 B. 144 C. 360 D. 720

C
The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
i.e, Required number of ways = $\frac {6!}{(1!)(2!)(1!)(1!)(1!)}$ = 360