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,$$ = 15 + 56 = 71 $$