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. |

- Find the number of ways arrangement of 12 people in 5 chairs?
- 92,120
- 95,040
- 97,023
- 98,258

Answer: (b) 95,040

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

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

- How many words can be formed from the letters of the word BANANA?
- 60
- 70
- 80
- 90

Answer: (a) 60

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

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

- How many 4 digit numbers can be formed by using 2, 3, 4, 5, 6, 7, when repetetion is not allowed?
- 220
- 280
- 360
- 380

Answer: (c) 360

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

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

- How many words can be formed from the letters of the word INDIA?
- 60
- 70
- 80
- 90

Answer: (a) 60 Words

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

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

- How many ways of sitting arrangement can be possible for 4 boys and 3 girls, if girls want to sit together?
- 560
- 660
- 700
- 720

Answer: (d) 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, $$ = 5! \times 3! $$ $$ = 120 \times 6 $$ $$ = 720 $$

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, $$ = 5! \times 3! $$ $$ = 120 \times 6 $$ $$ = 720 $$

- How many ways of sitting arrangement can be possible for 4 boys and 3 girls, if no two girls sit together?
- 1225
- 1440
- 1520
- 1650

Answer: (b) 1440

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, $$ = 5P_3 \times 4! $$ $$ = \frac{5!}{(5 - 3)!} \times 4! $$ $$ = \frac{5!}{2!} \times 4! $$ $$ = \frac{5 \times 4 \times 3 \times 2!}{2!} \times 24 $$ $$ = 1440 \ Ways $$

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, $$ = 5P_3 \times 4! $$ $$ = \frac{5!}{(5 - 3)!} \times 4! $$ $$ = \frac{5!}{2!} \times 4! $$ $$ = \frac{5 \times 4 \times 3 \times 2!}{2!} \times 24 $$ $$ = 1440 \ Ways $$

- How many ways 8 persons can be sit around a dinning table in 8 chairs?
- 5040
- 4152
- 4020
- 3822

Answer: (a) 5040

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.$$ = 7! $$ $$ = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 $$ $$ = 5040 \ Ways $$

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.$$ = 7! $$ $$ = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 $$ $$ = 5040 \ Ways $$

- How many three digit numbers can be formed by using 0, 1, 2, 3, 4, if repetetion is not allowed?
- 32
- 36
- 42
- 48

Answer: (d) 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, $$ = 4 \times 4 \times 3 $$ $$ = 48 $$

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, $$ = 4 \times 4 \times 3 $$ $$ = 48 $$

- Find the number of ways the arrangement of 9 people in 4 chairs?
- 3024
- 2856
- 2405
- 2425

Answer: (a) 3024

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

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

- How many words can be formed from the letters of the word ASIA?
- 8
- 10
- 12
- 14

Answer: (c) 12

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

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

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