# Permutation and Combination Quiz 14

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# Permutation and Combination Quiz 14

### Introduction

Permutation and Combination is one of important topic in Quantitative Aptitude Section. In Permutation and Combination Quiz 14 article candidates can find questions with answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Permutation and Combination Quiz 14 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

In how many different ways can the letters of the word "CHARGES" be arranged in such a way that the vowels always come together?
A. 1440 B. 720 C. 360 D. 240

A
The arrangement is made in such a way that the vowels always come together.
i.e., "CHRGS(AE)".
Considering vowels as one letter, 6 different letters can be arranged in 6! ways; i.e., 6! = 720 ways.
The vowels "AE" can be arranged themselves in 2! ways; i.e.,2! = 2 ways
Therefore, required number of ways = 720 x 2 = 1440 ways.

### Q2

In how many different ways can the letters of the word "COMPLAINT" be arranged in such a way that the vowels occupy only the odd positions?
A. 1440 B. 43200 C. 1440 D. 5420

B
There are 9 different letters in the given word "COMPLAINT", out of which there are 3 vowels and 6 consonants.
Let us mark these positions as under:
[1] [2] [3] [4] [5] [6] [7] [8] [9]
Now, 3 vowels can be placed at any of the three places out of 5 marked 1, 3, 5, 7 and 9.
Number of ways of arranging the vowels = $^{5}{P}_{3}$ = 5 x 4 x 3 = 60 ways.
Also, the 6 consonants at the remaining positions may be arranged in $^{6}{P}_{6}$ ways = 6! ways = 720 ways.
Therefore, required number of ways = 60 x 720 = 43200 ways.

### Q3

In how many different ways can the letters of the word "CANDIDATE" be arranged in such a way that the vowels always come together?
A. 4320 B. 1440 C. 720 D. 840

A
There are 9 letters in the given word, out of which 4 are vowels.
In the word "CANDIDATE" we treat the vowels "AIAE" as one letter.
Thus, we have CNDDT(AIAE).
Now, we have to arrange 6 letters, out of which D occurs twice.
Therefore, number of ways of arranging these letters = $\frac {6!}{2!} = \frac {720}{2} = 360 ways.$
Now, AIAE has 4 letters, in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = $\frac {4!}{2!} = \frac {1 \times 2 \times 3 \times 4 }{2} = 12$
Therefore, required number of words = (360 x 12) = 4320.

### Q4

In how many different ways can the letters of the word "RADIUS" be arranged in such a way that the vowels occupy only the odd positions?
A. 72 B. 144 C. 532 D. 36

D
There are 6 different letters in the given word, out of which there are 3 vowels and 3 consonants.
Let us mark these positions as under:
[1] [2] [3] [4] [5] [6]
Now, 3 vowels can be placed at any of the three places out of 3 marked 1, 3 and 5.
Number of ways of arranging the vowels = $^{3}{P}_{3}$ = 3! = 6 ways.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = $^{3}{P}_{3} = 3! = 6$ ways.
Therefore, total number of ways = 6 x 6 = 36.

### Q5

In how many different ways can the letters of the word 'SERVING' be arranged?
A. 5040 B. 720 C. 120 D. 40320

A
There are 7 different letters in the word 'SERVING'.
Therefore, the number of arrangements of the seven letters of the word = Number of all permutations of 7 letters, taken 7 at a time = $^{n}{P}_{n}$ = n(n - 1)(n - 2) ... (n - n + 1) = n!
Here, n = 7 then required number of ways = 7! = 5040.