# Permutation and Combination Aptitude Questions and Answers:

#### Overview:

 Questions and Answers Type: MCQ (Multiple Choice Questions). Main Topic: Quantitative Aptitude. Quantitative Aptitude Sub-topic: Permutation and Combination Aptitude Questions and Answers. Number of Questions: 10 Questions with Solutions.

1. Find the number of ways arrangement of 12 people in 5 chairs?

1. 92,120
2. 95,040
3. 97,023
4. 98,258

Solution: $$nP_r = \frac{n!}{(n - r)!}$$ $$12P_5 = \frac{12!}{(12 - 5)!}$$ $$= \frac{12!}{7!}$$ $$= \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{7!}$$ $$= 95,040$$

1. How many words can be formed from the letters of the word BANANA?

1. 60
2. 70
3. 80
4. 90

Solution: The word BANANA contains 3 A's, 2 N's, and 1 B, so the number of words can be formed.$$= \frac{6!}{3! \times 2!}$$ $$= \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1 \times 2 \times 1}$$ $$= 60 \ Words$$

1. How many 4 digit numbers can be formed by using 2, 3, 4, 5, 6, 7, when repetetion is not allowed?

1. 220
2. 280
3. 360
4. 380

Solution: Four digit numbers can be formed in $$6P_4$$ ways.$$6P_4 = \frac{6!}{(6 - 4)!}$$ $$= \frac{6!}{2!}$$ $$= \frac{6 \times 5 \times 4 \times 3 \times 2!}{2!}$$ $$= 360$$

1. How many words can be formed from the letters of the word INDIA?

1. 60
2. 70
3. 80
4. 90

Solution: The word INDIA contains 2 I's and one each A, D, and N, so the number of words can be formed, $$= \frac{5!}{2!}$$ $$= \frac{5 \times 4 \times 3 \times 2!}{2!}$$ $$= 60 \ Words$$

1. How many ways of sitting arrangement can be possible for 4 boys and 3 girls, if girls want to sit together?

1. 560
2. 660
3. 700
4. 720

Solution: $$B_1 B_2 B_3 B_4 G_1 G_2 G_3$$ considering all girls K, because they are sitting together, so the arrangement will be, $$B_1 B_2 B_3 B_4 K$$ it can be arranged in $$5!$$ ways.

Now, 3 girls can be arranged in $$3!$$ ways within their group.

So, final arrangement will be, $$= 5! \times 3!$$ $$= 120 \times 6$$ $$= 720$$

1. How many ways of sitting arrangement can be possible for 4 boys and 3 girls, if no two girls sit together?

1. 1225
2. 1440
3. 1520
4. 1650

Solution: The girls can be sit around the boys like this, _ $$B_1$$ _ $$B_2$$ _ $$B_3$$ _ $$B_4$$ _.

The boys can be arranged in $$4!$$ ways.

The girls can be arranged in vacant seats around the boys in $$5P_3$$ ways.

Hence, final arrangement can be done, $$= 5P_3 \times 4!$$ $$= \frac{5!}{(5 - 3)!} \times 4!$$ $$= \frac{5!}{2!} \times 4!$$ $$= \frac{5 \times 4 \times 3 \times 2!}{2!} \times 24$$ $$= 1440 \ Ways$$

1. How many ways 8 persons can be sit around a dinning table in 8 chairs?

1. 5040
2. 4152
3. 4020
4. 3822

Solution: For circular permutation total number of ways of arrangement = $$(n - 1)!$$

So, here 8 persons can be sit around a dinning table in 8 chairs by $$(8 - 1)!$$ ways.$$= 7!$$ $$= 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$$ $$= 5040 \ Ways$$

1. How many three digit numbers can be formed by using 0, 1, 2, 3, 4, if repetetion is not allowed?

1. 32
2. 36
3. 42
4. 48

Solution: Zero can not come in the first place, only 4 numbers (1, 2, 3, 4) can be come in first place. So rest of the numbers can be come in next places. The three digit numbers can be formed, $$= 4 \times 4 \times 3$$ $$= 48$$

1. Find the number of ways the arrangement of 9 people in 4 chairs?

1. 3024
2. 2856
3. 2405
4. 2425

Solution: $$nP_r = \frac{n!}{(n - r)!}$$ $$9P_4 = \frac{9!}{(9 - 4)!}$$ $$= \frac{9!}{5!}$$ $$= 3024 \ Ways$$

1. How many words can be formed from the letters of the word ASIA?

1. 8
2. 10
3. 12
4. 14

Solution: The word ASIA contains 2 A's and each one S, and I, so the number of words can be formed, $$= \frac{4!}{2!}$$ $$= 12 \ Words$$