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This mock test of Test: Permutation & Combination- 2 for UPSC helps you for every UPSC entrance exam.
This contains 10 Multiple Choice Questions for UPSC Test: Permutation & Combination- 2 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

MCQ (Multiple Choice Questions) with solution are available for Practice, which would help you prepare for Permutation and Combination under Quantitative Aptitude. You can practice these practice quizzes as per your speed and improvise the topic. The same topic is covered under various competitive examinations like - CAT, GMAT, Bank PO, SSC and other competitive examinations.

**Q. **Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Solution:

Number of groups, each having **3 consonants and 2 vowels** = 210. Required number of ways = (210 x 120) = 25200.

QUESTION: 2

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

Solution:

The total number of ways to select 4 children out of 10 total children is (104)=10∗9∗8∗74∗3∗2∗1=210. However, because at least one boy must be in the group, we have to subtract 1 to account for the case where all four children are girls (There is (44)=1 such case). The answer is 210−1=209.

QUESTION: 3

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?

Solution:

We have to consider the 3 possible ways to accomplish this :

5 men & 0 women

4 men & 1 women

3 men & 2 women

Hence the required number is

C(7,5).C(6,0) + C(7,4).C(6,1) + C(7,3).C(6,2)

=21x1 + 35x6 + 35x15

=21+210+525 = 756

QUESTION: 4

In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

Solution:

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. Required number of ways =(120∗6)=720.

QUESTION: 5

In how many different ways can the letters of the word 'MATHEMATICS' be arranged such that the vowels must always come together?

Solution:

In the word 'MATHEMATICS', we'll consider all the vowels AEAI together as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice

Number of ways of arranging these letters =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 =4! / 2!= 12.

Required number of words = (10080 x 12) = **120960**

QUESTION: 6

In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?

Solution:

Vowels in the word "CORPORATION" are O,O,A,I,O

Lets make it as CRPRTN(OOAIO)

This has 7 lettes, where R is twice so value = 7!/2!

= 2520

Vowel O is 3 times, so vowels can be arranged = 5!/3!

= 20

Total number of words = 2520 * 20 = 50400

QUESTION: 7

In how many different ways can the letters of the word 'LEADING' be arranged such that the vowels should always come together?

Solution:

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) 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.

QUESTION: 8

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?

Solution:

5 men can be chosen out of 7 men in ^{7}C_{5} = ^{7}C_{2} = 21 ways.

2 women out of 3 women in ^{3}C_{2} = ^{3}C_{1} = 3 ways.

Hence by fundamental counting principle, group of 5 men and 2 women can be chosen from 7 men and 3 women in (21)(3) = 63 ways

QUESTION: 9

How many 3-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

Solution:

The word 'LOGARITHMS' has 10 different letters.

Hence, the number of 3-letter words(with or without meaning) formed by using these letters

= ^{10}P_{3}

= 10 x 9 x 8

= 720

QUESTION: 10

There are 8 men and 10 women and you need to form a committee of 5 men and 6 women. In how many ways can the committee be formed?

Solution:

Required number of ways= (^{8}C_{5} x ^{10}C_{6})

= (^{8}C_{3} x ^{10}C_{4})

= (8 x 7 x 6 /3 x 2 x 1) x (10 x 9 x 8 x 7 / 4 x 3 x 2 x 1)

= 11760.

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