# 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. Five students are participating in a competition. In how many ways can the first two prizes be won?

1. 20
2. 25
3. 30
4. 35

Answer: (a) 20

Solution: The first two prizes can be won by $$5P_2$$ ways.$$5P_2 = \frac{5!}{(5 - 2)!}$$ $$= \frac{5!}{3!}$$ $$= 20 \ Ways$$

1. $$\sum_{r=1}^4 C(4, r) = ?$$

1. 12
2. 15
3. 18
4. 20

Answer: (b) 15

Solution: By using Binomial theorem,

$$nC_0 + nC_1 + nC_2 + ....nC_n$$ = $$2^n$$

$$nC_1 + nC_2 + ....nC_n$$ = $$2^n - nC_0$$

$$\sum_{r=1}^4 C(4, r)$$ = $$4C_1 + 4C_2 + 4C_3 + 4C_4$$

$$= 2^4 - 4C_0$$

$$= 16 - 1 = 15$$

1. In how many ways, out of 7 Black and 5 White balls, 4 Black and 3 White balls can be drawn?

1. 325
2. 350
3. 375
4. 380

Answer: (b) 350

Solution: The balls can be drawn in $$7C_4 \times 5C_3$$ ways.$$= 7C_4 \times 5C_3$$ $$= \frac{7!}{(7 - 4)! \times 4!} \times \frac{5!}{(5 - 3)! \times 3!}$$ $$= \frac{7!}{3! \times 4!} \times \frac{5!}{2! \times 3!}$$ $$= 350 \ Ways$$

1. If there are five points lie on a circle, then how many chords can be drawn by joning these points?

1. 10
2. 12
3. 15
4. 16

Answer: (a) 10

Solution: Number of chords can be drawn by $$5C_2$$ ways.$$5C_2 = \frac{5!}{(5 - 2)! \times 2!}$$ $$= \frac{5!}{3! \times 2!}$$ $$= 10 \ Chords$$

1. Six people A, B, C, D, E, and F occupy seats in a row such that C and D sit next to each other. In how many ways can these six people can sit?

1. 180
2. 212
3. 226
4. 240

Answer: (d) 240

Solution: If we assume C and D as one, then these people can sit by $$5!$$ ways, but C and D can also sit 2 ways within their group. Hence, $$= 5! \times 2$$ $$= 240 \ Ways$$

1. Find the value of $$8C_5 + 7C_4 + 6C_3$$

1. 111
2. 112
3. 113
4. 110

Answer: (a) 111

Solution: $$8C_5 = \frac{8!}{(8 - 5)! \times 5!}$$ $$= \frac{8!}{3! \times 5!}$$ $$= 56$$ $$7C_4 = \frac{7!}{(7 - 4)! \times 4!}$$ $$= \frac{7!}{3! \times 4!}$$ $$= 35$$ $$6C_3 = \frac{6!}{(6 - 3)! \times 3!}$$ $$= \frac{6!}{3! \times 3!}$$ $$= 20$$ Hence,$$= 56 + 35 + 20$$ $$= 111$$

1. In how many different ways can the letters of the word AUSTRALIA be arranged such that the vowels always come together?

1. 2200
2. 2400
3. 2500
4. 2600

Answer: (b) 2400

Solution: The word AUSTRALIA contains 5 vowels come together, so treating all vowels as one, it can be arranged in $$5!$$ ways. And the vowels (AUAIA) contains 3 A's within their group, so the vowels can be arranged within their group in $$\frac{5!}{3!}$$ ways. So, the final arrangement,$$= 5! \times \frac{5!}{3!}$$ $$= 2400 \ Ways$$

1. Seven students are participating in a competition. In how many ways can the first two prizes be won?

1. 201
2. 205
3. 210
4. 220

Answer: (c) 210

Solution: The first three prizes can be won by $$7P_3$$ ways.$$7P_3 = \frac{7!}{(7 - 3)!}$$ $$= \frac{7!}{4!}$$ $$= 210 \ Ways$$

1. In how many ways, out of 9 Red and 7 Yellow balls, 4 Red and 3 Yellow balls can be drawn?

1. 4410
2. 4420
3. 4425
4. 4430

Answer: (a) 4410

Solution: The balls can be drawn in $$9C_4 \times 7C_3$$ ways.$$9C_4 = \frac{9!}{(9 - 4)! \times 4!}$$ $$= \frac{9!}{5! \times 4!}$$ $$= 126$$ $$7C_3 = \frac{7!}{(7 - 3)! \times 3!}$$ $$= \frac{7!}{4! \times 3!}$$ $$= 35$$ Hence,$$= 126 \times 35$$ $$= 4410 \ Ways$$

1. Find the value of $$6C_4 + 8C_5$$

1. 56
2. 63
3. 69
4. 71

Answer: (d) 71

Solution: $$6C_4 = \frac{6!}{(6 - 4)! \times 4!}$$ $$= \frac{6!}{2! \times 4!}$$ $$= 15$$ $$8C_5 = \frac{8!}{(8 - 5)! \times 5!}$$ $$= \frac{8!}{3! \times 5!}$$ $$= 56$$ Hence,$$= 15 + 56 = 71$$