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 10 people in 7 chairs?
- 502,600
- 604,800
- 705,500
- 825,200

Answer: (b) 604,800

Solution: $$ nP_r = \frac{n!}{(n - r)!} $$ $$ 10P_7 = \frac{10!}{(10 - 7)!} $$ $$ = \frac{10!}{3!} $$ $$ = 604,800 $$

Solution: $$ nP_r = \frac{n!}{(n - r)!} $$ $$ 10P_7 = \frac{10!}{(10 - 7)!} $$ $$ = \frac{10!}{3!} $$ $$ = 604,800 $$

- How many words can be formed from the letters of the word AMERICA?
- 2,520
- 2,846
- 3,225
- 3,232

Answer: (a) 2,520

Solution: The word BANANA contains 2 A's, and each one M, E, R, I, and C, so the number of words can be formed.$$ = \frac{7!}{2!} $$ $$ = 2,520 \ Words $$

Solution: The word BANANA contains 2 A's, and each one M, E, R, I, and C, so the number of words can be formed.$$ = \frac{7!}{2!} $$ $$ = 2,520 \ Words $$

- How many ways of sitting arrangement can be possible for 5 Men and 4 Women, if Women want to sit together?
- 14,265
- 15,212
- 16,250
- 17,280

Answer: (d) 17,280

Solution: $$ M_1 M_2 M_3 M_4 M_5 W_1 W_2 W_3 W_4 $$ considering all women K, because they are sitting together, so the arrangement will be, $$ M_1 M_2 M_3 M_4 M_5 K $$ it can be arranged in \(6!\) ways.

Now, 4 women can be arranged in \(4!\) ways within their group.

So, final arrangement will be, $$ = 6! \times 4! $$ $$ = 720 \times 24 $$ $$ = 17,280 $$

Solution: $$ M_1 M_2 M_3 M_4 M_5 W_1 W_2 W_3 W_4 $$ considering all women K, because they are sitting together, so the arrangement will be, $$ M_1 M_2 M_3 M_4 M_5 K $$ it can be arranged in \(6!\) ways.

Now, 4 women can be arranged in \(4!\) ways within their group.

So, final arrangement will be, $$ = 6! \times 4! $$ $$ = 720 \times 24 $$ $$ = 17,280 $$

- How many ways of sitting arrangement can be possible for 5 Men and 4 Women, if no two Women sit together?
- 43,200
- 45,300
- 47,300
- 48,500

Answer: (a) 43,200

Solution: The women can be sit around the men like this, _ \(M_1\) _ \(M_2\) _ \(M_3\) _ \(M_4\) _ \(M_5\) _.

The men can be arranged in \(5!\) ways.

The women can be arranged in vacant seats around the boys in \(6P_4\) ways.

Hence, final arrangement can be done, $$ = 6P_4 \times 5! $$ $$ = \frac{6!}{(6 - 4)!} \times 5! $$ $$ = \frac{6!}{2!} \times 5! $$ $$ = 43,200 \ Ways $$

Solution: The women can be sit around the men like this, _ \(M_1\) _ \(M_2\) _ \(M_3\) _ \(M_4\) _ \(M_5\) _.

The men can be arranged in \(5!\) ways.

The women can be arranged in vacant seats around the boys in \(6P_4\) ways.

Hence, final arrangement can be done, $$ = 6P_4 \times 5! $$ $$ = \frac{6!}{(6 - 4)!} \times 5! $$ $$ = \frac{6!}{2!} \times 5! $$ $$ = 43,200 \ Ways $$

- How many four digit numbers can be formed by using 0, 1, 2, 3, 4, if repetetion is not allowed?
- 56
- 66
- 86
- 96

Answer: (d) 96

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 four digit numbers can be formed, $$ = 4 \times 4 \times 3 \times 2 $$ $$ = 96 $$

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 four digit numbers can be formed, $$ = 4 \times 4 \times 3 \times 2 $$ $$ = 96 $$

- How many new words can be formed by using the letters of the word PHILIPPINES?
- 1,108,799
- 1,108,800
- 1,108,850
- 1,108,859

Answer: (a) 1,108,799

Solution: Here, we need to form new words, but the word PHILIPPINES is already a word, so we will not count the word PHILIPPINES as a new word. And the word PHILIPPINES contains 3 P's, 3 I's, and each one H, L, N, E, and S. Hence, new words can be formed, $$ = \frac{11!}{3! \times 3!} - 1 $$ $$ = 1,108,800 - 1 $$ $$ = 1,108,799 \ words $$

Solution: Here, we need to form new words, but the word PHILIPPINES is already a word, so we will not count the word PHILIPPINES as a new word. And the word PHILIPPINES contains 3 P's, 3 I's, and each one H, L, N, E, and S. Hence, new words can be formed, $$ = \frac{11!}{3! \times 3!} - 1 $$ $$ = 1,108,800 - 1 $$ $$ = 1,108,799 \ words $$

- In how many ways can the letters of the word QUALITY be arranged so that all the vowels come together?
- 280
- 320
- 360
- 380

Answer: (c) 360

Solution: The word QUALITY contains 3 vowels A, I, U, and 4 consonents Q, L, T, Y.

If we consider all 3 vowels as one unit (X), then it can be arranged by 5! ways.

Q, L, T, Y, X

and the vowels itself can be arranged by \(3!\) ways. Hence, final arrangement, $$ = 5! \times 3! $$ $$ = 360 \ ways $$

Solution: The word QUALITY contains 3 vowels A, I, U, and 4 consonents Q, L, T, Y.

If we consider all 3 vowels as one unit (X), then it can be arranged by 5! ways.

Q, L, T, Y, X

and the vowels itself can be arranged by \(3!\) ways. Hence, final arrangement, $$ = 5! \times 3! $$ $$ = 360 \ ways $$

- How many words can be formed by using the letters of the word STUDENT, if words start with the letter S?
- 360
- 320
- 252
- 236

Answer: (a) 360

Solution: In the word STUDENT, Here S is fixed, so we have all the letters to arrange except 'S'. So it can be arranged, $$ = \frac{6!}{2!} $$ $$ = 360 \ words $$

Solution: In the word STUDENT, Here S is fixed, so we have all the letters to arrange except 'S'. So it can be arranged, $$ = \frac{6!}{2!} $$ $$ = 360 \ words $$

- How many ways are there to house 6 persons in a home, if there are 3 rooms in the home, one single, one double, and one for three persons?
- 50
- 60
- 70
- 80

Answer: (b) 60

Solution: We can house 6 persons, $$ = \frac{6!}{1! \times 2! \times 3!} $$ $$ = 60 \ ways $$

Solution: We can house 6 persons, $$ = \frac{6!}{1! \times 2! \times 3!} $$ $$ = 60 \ ways $$

- In how many ways can 4 cricket players be selected from a group of 10?
- 180
- 199
- 210
- 219

Answer: (c) 210

Solution: 4 cricket players can be selected from the group of 10 by \(10C_4\) ways.

Hence, $$ 10C_4 = \frac{10!}{(10 - 4)! \times 4!} $$ $$ = \frac{10!}{6! \times 4!} $$ $$ = 210 \ Ways $$

Solution: 4 cricket players can be selected from the group of 10 by \(10C_4\) ways.

Hence, $$ 10C_4 = \frac{10!}{(10 - 4)! \times 4!} $$ $$ = \frac{10!}{6! \times 4!} $$ $$ = 210 \ Ways $$

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