Permutation and Combination – Quiz 6

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Permutation and Combination – Quiz 6

Introduction

Permutation and Combination is one of important topic in Quantitative Aptitude Section. In Permutation and Combination – Quiz 6 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 6 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 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?
A. 32 B. 40 C. 36 D. 60

C
There are 6 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 4, marked 1, 3, 5.
Number of ways of arranging the vowels = $^{3}{P}_{3}$ = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = $^{3}{P}_{3}$ = 3! = 6.
Total number of ways = (6 x 6) = 36

Q2

In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
A. 63 B. 90 C. 126 D. 45

A
= $(^{3}{P}_{3} \times ^{3}{P}_{3}) = (^{7}{C}_{2} \times ^{3}{C}_{1}) = (\frac {7 \times 6}{2 \times 1} \times 3) = 63$

Q3

How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?
A. 40 B. 400 C. 5040 D. 2520

C
'LOGARITHMS' contains 10 different letters.
Required number of words = Number of arrangements of 10 letters, taking 4 at a time.
= $^{10}{P}_{4}$
= (10 x 9 x 8 x 7)
= 5040.

Q4

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?
A. 10080 B. 4989600 C. 120960 D. None of these

C
In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
i.e, Number of ways of arranging these letters = $\frac {8!}{(2!)(2!)}$ = 10080.
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = $\frac {4!}{2!}$ = 12
i.e, Required number of words = (10080 x 12) = 120960.

Q5

In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
A. 120 B. 720 C. 4320 D. 2160

B
The word 'OPTICAL' contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
i.e, Required number of ways = (120 x 6) = 720.

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