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. |

- Five students are participating in a competition. In how many ways can the first two prizes be won?
- 20
- 25
- 30
- 35

Answer: (a) 20

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

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

- \(\sum_{r=1}^4 C(4, r) = ?\)
- 12
- 15
- 18
- 20

Answer: (b) 15

Solution: By using Binomial theorem,

\(nC_0 + nC_1 + nC_2 + ....nC_n\) = \(2^n\)

\(nC_1 + nC_2 + ....nC_n\) = \(2^n - nC_0\)

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

\(= 2^4 - 4C_0\)

\(= 16 - 1 = 15\)

Solution: By using Binomial theorem,

\(nC_0 + nC_1 + nC_2 + ....nC_n\) = \(2^n\)

\(nC_1 + nC_2 + ....nC_n\) = \(2^n - nC_0\)

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

\(= 2^4 - 4C_0\)

\(= 16 - 1 = 15\)

- In how many ways, out of 7 Black and 5 White balls, 4 Black and 3 White balls can be drawn?
- 325
- 350
- 375
- 380

Answer: (b) 350

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

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

- If there are five points lie on a circle, then how many chords can be drawn by joning these points?
- 10
- 12
- 15
- 16

Answer: (a) 10

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

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

- 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?
- 180
- 212
- 226
- 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, $$ = 5! \times 2 $$ $$ = 240 \ Ways $$

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, $$ = 5! \times 2 $$ $$ = 240 \ Ways $$

- Find the value of \(8C_5 + 7C_4 + 6C_3\)
- 111
- 112
- 113
- 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,$$ = 56 + 35 + 20 $$ $$ = 111 $$

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,$$ = 56 + 35 + 20 $$ $$ = 111 $$

- In how many different ways can the letters of the word AUSTRALIA be arranged such that the vowels always come together?
- 2200
- 2400
- 2500
- 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,$$ = 5! \times \frac{5!}{3!} $$ $$ = 2400 \ Ways $$

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,$$ = 5! \times \frac{5!}{3!} $$ $$ = 2400 \ Ways $$

- Seven students are participating in a competition. In how many ways can the first two prizes be won?
- 201
- 205
- 210
- 220

Answer: (c) 210

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

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

- In how many ways, out of 9 Red and 7 Yellow balls, 4 Red and 3 Yellow balls can be drawn?
- 4410
- 4420
- 4425
- 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,$$ = 126 \times 35 $$ $$ = 4410 \ Ways $$

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,$$ = 126 \times 35 $$ $$ = 4410 \ Ways $$

- Find the value of \(6C_4 + 8C_5\)
- 56
- 63
- 69
- 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,$$ = 15 + 56 = 71 $$

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,$$ = 15 + 56 = 71 $$

Lec 1: Permutation and Combination
Lec 2: Binomial Theorem
Exercise-1
Exercise-2
Exercise-3
Exercise-4
Exercise-5