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. |

- How many numbers of 4 digits can be formed with the digits 1, 2, 3, 4, 5, 6, 7, when the repetition of digits not allowed?
- 890
- 870
- 840
- 810

Answer: (c) 840

Solution: The 4 digit numbers can be formed in \(7P_4\) ways.$$ 7P_4 = \frac{7!}{(7 - 4)!} $$ $$ = \frac{7!}{3!} $$ $$ = 840 $$

Solution: The 4 digit numbers can be formed in \(7P_4\) ways.$$ 7P_4 = \frac{7!}{(7 - 4)!} $$ $$ = \frac{7!}{3!} $$ $$ = 840 $$

- In how many ways 5 Indians, 4 Russians, and 2 Americans be seated in a row so that all persons of the same nationality sit together?
- 34,560
- 32,376
- 28,660
- 26,852

Answer: (a) 34,560

Solution: Indians can sit in \(5!\) ways.

Russians can sit in \(4!\) ways.

Americans can sit in \(2!\) ways.

And all of them can sit together in \(3!\) different ways. Hence, final arrangement,$$ = 5! \times 4! \times 2! \times 3! $$ $$ = 34,560 \ ways $$

Solution: Indians can sit in \(5!\) ways.

Russians can sit in \(4!\) ways.

Americans can sit in \(2!\) ways.

And all of them can sit together in \(3!\) different ways. Hence, final arrangement,$$ = 5! \times 4! \times 2! \times 3! $$ $$ = 34,560 \ ways $$

- In how many ways the letters of the word COLLEGE can be arranged?
- 1850
- 1520
- 1466
- 1260

Answer: (d) 1260

Solution: The word COLLEGE contains 2 L's, 2 E's and one each C, O, and G, so it can be arranged,$$ = \frac{7!}{2! \times 2!} $$ $$ = 1260 \ ways $$

Solution: The word COLLEGE contains 2 L's, 2 E's and one each C, O, and G, so it can be arranged,$$ = \frac{7!}{2! \times 2!} $$ $$ = 1260 \ ways $$

- How many words can be formed by using the letters of the word COLLEGE, if words start with the letter C?
- 120
- 160
- 180
- 220

Answer: (c) 180

Solution: In the word COLLEGE, Here C is fixed, so we have all the letters to arrange except 'C'. So it can be arranged, $$ = \frac{6!}{2! \times 2!} $$ $$ = 180 \ words $$

Solution: In the word COLLEGE, Here C is fixed, so we have all the letters to arrange except 'C'. So it can be arranged, $$ = \frac{6!}{2! \times 2!} $$ $$ = 180 \ words $$

- In how many ways, a captain, a vice-captain, and a wicket-keeper can be chosen from the 11 players of a cricket team?
- 990
- 860
- 750
- 660

Answer: (a) 990

Solution: 3 positions can be chosen from 11 players by \(11C_3\) ways and these 3 positions can be chosen within the 3 players by \(3!\) ways. So,$$ = 11C_3 \times 3! $$ $$ = \frac{11!}{(11 - 3)! \times 3!} \times 3! $$ $$ = \frac{11!}{8!} $$ $$ = 990 \ ways $$

Solution: 3 positions can be chosen from 11 players by \(11C_3\) ways and these 3 positions can be chosen within the 3 players by \(3!\) ways. So,$$ = 11C_3 \times 3! $$ $$ = \frac{11!}{(11 - 3)! \times 3!} \times 3! $$ $$ = \frac{11!}{8!} $$ $$ = 990 \ ways $$

- How many ways are there to house 10 persons in a hotel, if there are 4 rooms in a hotel, one single, one double, one for three persons, and one for four persons?
- 10,300
- 12,600
- 13,400
- 14,200

Answer: (b) 12,600

Solution: We can house 10 persons, $$ = \frac{10!}{1! \times 2! \times 3! \times 4!} $$ $$ = 12,600 \ ways $$

Solution: We can house 10 persons, $$ = \frac{10!}{1! \times 2! \times 3! \times 4!} $$ $$ = 12,600 \ ways $$

- How many new words can be formed by using the letters of the word RUSSIA?
- 360
- 359
- 358
- 357

Answer: (b) 359

Solution: Here, we need to form new words, but the word RUSSIA is already a word, so we will not count the word RUSSIA as a new word. And the word RUSSIA contains 2 S's and each one R, U, I, and A. Hence, new words can be formed, $$ = \frac{6!}{2!} - 1 $$ $$ = 360 - 1 $$ $$ = 359 \ words $$

Solution: Here, we need to form new words, but the word RUSSIA is already a word, so we will not count the word RUSSIA as a new word. And the word RUSSIA contains 2 S's and each one R, U, I, and A. Hence, new words can be formed, $$ = \frac{6!}{2!} - 1 $$ $$ = 360 - 1 $$ $$ = 359 \ words $$

- In how many ways can the letters of the word EDUCATION be arranged so that all the vowels come together?
- 10,300
- 12,500
- 13,200
- 14,400

Answer: (d) 14,400

Solution: The word contains 5 vowels A, E, I, O, U, and 4 consonents D, C, T, N.

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

D, C, T, N, X

and the vowels itself can be arranged by \(5!\) ways. Hence, final arrangement, $$ = 5! \times 5! $$ $$ = 14,400 \ ways $$

Solution: The word contains 5 vowels A, E, I, O, U, and 4 consonents D, C, T, N.

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

D, C, T, N, X

and the vowels itself can be arranged by \(5!\) ways. Hence, final arrangement, $$ = 5! \times 5! $$ $$ = 14,400 \ ways $$

- How many numbers of 2 digits can be formed with the digits 4, 5, 6, 7, when the repetition of the digits not allowed?
- 12
- 10
- 8
- 6

Answer: (a) 12

Solution: The 2 digit numbers can be formed in \(4P_2\) ways.$$ 4P_2 = \frac{4!}{(4 - 2)!} $$ $$ = \frac{4!}{2!} $$ $$ = 12 \ ways $$

Solution: The 2 digit numbers can be formed in \(4P_2\) ways.$$ 4P_2 = \frac{4!}{(4 - 2)!} $$ $$ = \frac{4!}{2!} $$ $$ = 12 \ ways $$

- In how many ways the letters of the word UNIVERSITY can be arranged?
- \(\frac{10!}{2! \times 2!}\)
- \(\frac{10!}{2!}\)
- \(\frac{10!}{4! \times 2!}\)
- \(\frac{10!}{4!}\)

Answer: (b) \(\frac{10!}{2!}\)

Solution: The word UNIVERSITY contains 2 I's and each one U, N, V, E, R, S, T, Y, so it can be arranged.$$ = \frac{10!}{2!} \ ways $$

Solution: The word UNIVERSITY contains 2 I's and each one U, N, V, E, R, S, T, Y, so it can be arranged.$$ = \frac{10!}{2!} \ ways $$

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