## Quantitative Aptitude Questions and Answers - Permutation and Combination

11. In how many different ways can any 4 letters of the word 'ABOLISH' be arranged?

A. 5040

B. 120

C. 240

D. 840

View Answer

Ans : D

Explanation: There are 7 different letters in the word 'ABOLISH'.

Therefore,
The number of arrangements of any 4 out of seven letters of the word = Number of all permutations

of 7 letters, taken 4 at a time =

nPr = n(n - 1)(n - 2) ... (n - r + 1)

Here, n = 7 and r = 4, then we have

7p4 = 7 x 6 x 5 x 4 = 840.

Hence, the required number of ways is 840.

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

A. 1720

B. 4320

C. 2160

D. 2400

View Answer

Ans : B

Explanation: The word 'SPORADIC' contains 8 different letters.

When the vowels OAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters SPRDC (OAI).

Now, 6 letters can be arranged in 6! = 720ways.

The vowels (OAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (720 x 6) = 4320.

13. In how many different ways can the letters of the word 'DILUTE' be arranged such that the vowels may
appear in the even places?

A. 24

B. 120

C. 72

D. 36

View Answer

Ans : D

Explanation: There are 3 consonants and 3 vowels in the word DILUTE.

Out of 6 places, 3 places odd and 3 places are even.

3 vowels can arranged in 3 even places in 3p3 ways = 3! = 6 ways.

And then 3 consonants can be arranged in the remaining 3 places in 3p3 ways = 3! = 6 ways.

Hence, the required number of ways = 6 x 6 = 36

14. There are 7 periods in each working day of a college. In how many ways can one organize 6 subjects
such that each subject is allowed at least one period?

A. 33200

B. 15120

C. 10800

D. 43600

View Answer

Ans : B

Explanation: 6 subjects can be arranged in periods in 7P6 ways.

Remaining 1 period can be arranged in 6P1 ways.

Two subjects are alike in each of the arrangement.

So we need to divide by 2! to avoid over counting.

Total number of arrangements = (7P6 x
6P1)/2!

= 5040 × 6 / 2

= 30240 / 2

= 15120

15. Pramoth has 12 friends and he wants to invite 7 of them to a party. How many times will 4 particular
friends never attend the party?

A. 7

B. 8

C. 12

D. 15

View Answer

Ans : B

Explanation: Therefore , required number of ways = 8C7
= 8C1

= 8

16. A shopkeeper has 15 models of cup and 9 models of saucer. In how many ways can he make a pair of
cup and saucer?

A. 100

B. 80

C. 110

D. 135

View Answer

Ans : D

Explanation: He has 15 patterns of cup and 9 model of saucer

A cup can be selected in 15 ways.

A saucer can be selected in 9 ways.

Hence one cup and one saucer can be selected in 15×9 ways =135 ways

17. In a birthday party, every person shakes hand with every other person. If there was a total of 66
handshakes in the party, how many persons were present in the party?

A. 9

B. 10

C. 11

D. 12

View Answer

Ans : D

Explanation: Assume that in a party every person shakes hand with every other person

Number of hand shake = 66

Total number of hand shake is given by nC2

Let n = the total number of persons present in the party

nC2 =66

n (n-1) / 2 = 66

n^2 - n = 2 × 66

n^2 - n – 132= 0

n= 12 , -11

But we cannot take negative value of n

So, n = 12

Therefore number of persons in the party = 12

18. In how many ways can a team of 6 persons be formed out of a total of 12 persons such that 3 particular
persons should not be included in any team?

A. 56

B. 112

C. 84

D. 168

View Answer

Ans : C

Explanation: Three particular persons should not be included in each team.

i.e., we have to select remaining 6- 3= 3 persons from 12-3 = 9 persons.

Hence, required number of ways = 9C3

= {9×8 × 7} / {3 × 2 × 1}

= 504 / 6

= 84

19. How many 3-letter words can be formed with or without meaning from the letters A , G , M , D , N , and J
, which are ending with G and none of the letters should be repeated?

A. 18

B. 20

C. 25

D. 27

View Answer

Ans : B

Explanation: Since each desired word is ending with G, the least place is occupied with G.

So, there is only 1 way.

The second place can now be filled by any of the remaining 5 letters (A , M , D , N , J ). So, there are 5 ways
of filling that place.

Then, the first place can now be filled by any of the remaining 4 letters.

So, there are 4 ways to fill.
Required number of words = (1 x 5 x 4) = 20.

20. How many words can be formed by using all letters of the word CABIN?

A. 24

B. 60

C. 120

D. 720

View Answer

Ans : C

Explanation: The word 'CABIN' has 5 letters and all these 5 letters are different.

Total number of words that can be formed by using all these 5 letters

= 5P5

= 5!

= 5×4×3×2×1

= 120

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