## Permutation and Combination Aptitude Questions and Answers

This section focuses on "Permutation and Combination" of Quantitative Aptitude. These Multiple Choice Questions (mcq) should be practiced to improve the Quantitative Aptitude skills required for various interviews (campus interviews, walk-in interviews, company interviews), placements, entrance exams and other competitive examinations.

1. In how many ways can the letters of the word 'NOMINATION' be arranged?

A. 237672

B. 123144

C. 151200

D. 150720

View Answer

Ans : C

Explanation: The word 'NOMINATION' contains 10 letters, namely

3N, 2O, 1M, 2I,1A, and 1T

Required number of ways = 10 ! / (3!)(2!)(1!)(2!)(1!)(1!)

= 151200

2. In how many different ways can the letters of the word 'POTENCY' be arranged in such a way that the
vowels always come together?

A. 1140

B. 1240

C. 1340

D. 1440

View Answer

Ans : D

Explanation: The word 'POTENCY' has 7 different letters.

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

Then, we have to arrange the letters PTNCY (EO).

Now, 6 (5 + 1 = 6) letters can be arranged in 6! = 720 ways.

The vowels (EO) can be arranged among themselves in 2! = 2 ways.

Required number of ways = (720 x2)

= 1440.

3. In how many different ways can the letters of the word ""ZYMOGEN"" be arranged in such a way that the
vowels always come together?

A. 1440

B. 1720

C. 2360

D. 2240

View Answer

Ans : A

Explanation: The arrangement is made in such a way that the vowels always come together.

i.e., ""ZYMGN(OE)"".

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

4. Find out the number of ways in which 12 Bangles of different types can be worn in 2 hands?

A. 1260

B. 2720

C. 1225

D. 4096

View Answer

Ans : D

Explanation: The first bangle can be worn in any of the 2 hands (2 ways).

Similarly each of the remaining 11 bangles also can be worn in 2 ways.

Hence total number of ways=2×2×2×2×2×2×2×2×2×2×2×2

=2^12

=4096

5. In how many ways can a group of 10 men and 5 women be made out of a total of 12 men and 10 women?

A. 16632

B. 15290

C. 25126

D. 34845

View Answer

Ans : A

Explanation: Required number of ways = 12C10 x
10C5

= 66 × 252

= 16632

6. A bowl contains 8 violet, 6 purple and 4 magenta balls. Three balls are drawn at random. Find out the
number of ways of selecting the balls of different colours?

A. 362

B. 248

C. 122

D. 192

View Answer

Ans : D

Explanation: 1 violet ball can be selected is 8C1 ways.

1 purple ball can be selected in 6C1 ways.

1 magenta ball can be selected in 4C1 ways.

Total number of ways = 8C1 × 6C1 × 4C1

= 8×6×4

= 192

7. In how many ways can 6 girls be seated in a rectangular order?

A. 60

B. 120

C. 240

D. 720

View Answer

Ans : B

Explanation: Number of arrangements possible = (6-1)!

= 5!

= 5×4×3×2×1

= 120

8. In how many ways can 10 stones can be arranged to form a bangles

A. 267720

B. 125380

C. 181440

D. 284360

View Answer

Ans : C

Explanation: Number of arrangements possible = {1} / {2} X (10-1)!

= {1} / {2}X 9!

= {1} / {2} X 9×8×7× 6×5×4×3×2×1

= {1 / 2 } ×362880

= 181440

9. In how many ways can 8 different ballons be distributed among 7 different boxes when any box
can have any number of ballons?

A. 8^7-1

B. 7^8-1

C. 7^8

D. 8^7

View Answer

Ans : C

Explanation: Here n = 7, k = 8. Hence, required number of ways = n^k =7^8

10. How many different possible permutations can be made from the word ‘WAGGISH’ such that the vowels
are never together?

A. 3605

B. 3120

C. 1240

D. 1800

View Answer

Ans : D

Explanation: The word ‘WAGGISH’ contains 7 letters of which 1 letter occurs twice
= 7! / 2!

= 2520

No. of permutations possible with vowels always together = 6! * 2! / 2!

= 1440 / 2

= 720

No. of permutations possible with vowels never together = 2520-720

= 1800.

Discussion