Aptitude Questions on Permutation Combination

Review PnC Concepts

Q. 1 How many unique 5-letter words can be created using the letters of the word “APPLE” such that no letter is repeated?

A) 120

B) 60

C) 240

D) 30

Check Solution

Ans: B) 60

Since “APPLE” has 5 letters, with the letter ‘P’ repeating twice, the number of distinct 5-letter words that can be formed is given by: $\frac{5!}{2!} = \frac{120}{2} = 60$

Q. 2 You have 4 boys and 4 girls waiting to be seated in a row for a photograph. However, there’s a catch: no two girls are allowed to sit together. How many different ways can they be seated while maintaining this arrangement rule?

A) 576

B) 2880

C) 3456

D) 1728

Check Solution

Ans: B) 2880

Arrange 4 boys in 4! ways. This creates 5 gaps for the girls to sit in. We choose 4 of these 5 gaps for the girls in $\binom{5}{4}$ ways and arrange the girls in 4! ways.

$4! \times \binom{5}{4} \times 4! = 24 \times 5 \times 24 = 2880$

Q. 3 The word “INDEPENDENCE” is packed with repeating letters! If you were to arrange all its letters, how many unique ways could you come up with?

A) 453600

B) 378000

C) 1000000

D) 16800

Check Solution

Ans: A) 453600

The word “INDEPENDENCE” has 12 letters: I, N (3 times), D (2 times), E (4 times), and C. So,

$\frac{12!}{3! \times 2! \times 4! \times 1!} = \frac{479001600}{6 \times 2 \times 24 \times 1} = 453600$

Q. 4 You have 10 identical balls and 4 unique boxes in front of you. How many ways can you distribute all the balls among the boxes? (Hint: Since the balls are identical, it’s the distribution that matters, not the arrangement)

A) 220

B) 84

C) 120

D) 286

Check Solution

Ans: D) 286

This is a problem of distributing n identical items into r distinct groups, which is given by $\binom{n + r – 1}{r – 1}$.

$\binom{10 + 4 – 1}{4 – 1} = \binom{13}{3} = 286$

Q. 5 From a group of 8 men and 7 women, a committee of 6 members is to be formed. The only condition is that the committee must have at least 3 women. How many ways can this committee be chosen while meeting the requirement?

A) 4000

B) 4100

C) 3100

D) 3108

Check Solution

Ans: D) 3108

Case 1: 3 women, 3 men: (73)×(83)=$\binom{7}{3} \times \binom{8}{3} = 35 \times 56 = 1960$
Case 2: 4 women, 2 men: (74)×(82)=$\binom{7}{4} \times \binom{8}{2} = 35 \times 28 = 980$
Case 3: 5 women, 1 man: (75)×(81)=$\binom{7}{5} \times \binom{8}{1} = 21 \times 8 = 168$

$1960+980+168=3108$

Q. 6 The word “MISSISSIPPI” has so many repeating letters! How many unique arrangements can you make if all the S’s must always stick together as a single block? (Hint: It’s like treating the S’s as one super letter—how does this change your possibilities?)

A) 18000

B) 15000

C) 420

D) 1680

Check Solution

Ans: C) 420

Treat all S’s as one unit. We then have: M, I, I, I, P, P, S block. The total number of arrangements is:

$\frac{7!}{3! \times 2!} = \frac{5040}{6 \times 2} = 420$

Q. 7 You need to form a committee of 4 members from a group of 10 employees. However, a stubborn couple among the employees insist on being included in every committee. With this condition in place, how many ways can you choose the remaining members to complete the group?

A) 28

B) 56

C) 8

D) 20

Check Solution

Ans: A) 28

If 2 particular people (the couple) are always included, we need to choose 2 more people from the remaining 8. This can be done in:

$\binom{8}{2} = 28$

Q. 8 Using the letters of the word “ARRANGEMENT,” how many distinct arrangements can you make if all the vowels must always appear together as a single unit?

A) 43200

B) 60480

C) 75600

D) 50400

Check Solution

Ans: B) 60480

Treat the vowels (A, A, E, E) as a single unit, giving us 8 units. Then,

$\frac{8!}{2! \times 2!} \times \frac{4!}{2! \times 2!} = 10080 \times 6 = 60480$

Q. 9 An international committee of 5 people is to be formed from an elected group of 7 men and 6 women. However, the requirement is that at least 2 women must be included in the group to represent the diversity. How many different ways can this be achieved while adhering to the condition?

A) 756

B) 630

C) 850

D) 1050

Check Solution

Ans: D) 1050

Consider cases for the number of women:

• 2 women, 3 men: $\binom{6}{2} \times \binom{7}{3} = 15 \times 35 = 525$
• 3 women, 2 men: $\binom{6}{3} \times \binom{7}{2} = 20 \times 21 = 420$
• 4 women, 1 man: $\binom{6}{4} \times \binom{7}{1} = 15 \times 7 = 105$

Total: $525+420+105=1050$

Q. 10 Imagine you have 5 identical apples and 3 identical oranges, and you want to distribute them among 4 children. How many different ways can this distribution be carried out? (Hint: It’s the distribution that counts, not who gets which specific fruit.)

A) 210

B) 1120

C) 120

D) 56

Check Solution

Ans: B) 1120

This is a problem of distributing identical items among distinct groups with a total number of ways:

$\binom{5+4-1}{4-1} \times \binom{3+4-1}{4-1} = \binom{8}{3} \times \binom{6}{3} = 56 \times 20 = 1120 $

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