# 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. 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?

1. 890
2. 870
3. 840
4. 810

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

1. 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?

1. 34,560
2. 32,376
3. 28,660
4. 26,852

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,$$= 5! \times 4! \times 2! \times 3!$$ $$= 34,560 \ ways$$

1. In how many ways the letters of the word COLLEGE can be arranged?

1. 1850
2. 1520
3. 1466
4. 1260

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

1. How many words can be formed by using the letters of the word COLLEGE, if words start with the letter C?

1. 120
2. 160
3. 180
4. 220

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

1. In how many ways, a captain, a vice-captain, and a wicket-keeper can be chosen from the 11 players of a cricket team?

1. 990
2. 860
3. 750
4. 660

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,$$= 11C_3 \times 3!$$ $$= \frac{11!}{(11 - 3)! \times 3!} \times 3!$$ $$= \frac{11!}{8!}$$ $$= 990 \ ways$$

1. 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?

1. 10,300
2. 12,600
3. 13,400
4. 14,200

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

1. How many new words can be formed by using the letters of the word RUSSIA?

1. 360
2. 359
3. 358
4. 357

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, $$= \frac{6!}{2!} - 1$$ $$= 360 - 1$$ $$= 359 \ words$$

1. In how many ways can the letters of the word EDUCATION be arranged so that all the vowels come together?

1. 10,300
2. 12,500
3. 13,200
4. 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, $$= 5! \times 5!$$ $$= 14,400 \ ways$$

1. How many numbers of 2 digits can be formed with the digits 4, 5, 6, 7, when the repetition of the digits not allowed?

1. 12
2. 10
3. 8
4. 6

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

1. In how many ways the letters of the word UNIVERSITY can be arranged?

1. $$\frac{10!}{2! \times 2!}$$
2. $$\frac{10!}{2!}$$
3. $$\frac{10!}{4! \times 2!}$$
4. $$\frac{10!}{4!}$$

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.$$= \frac{10!}{2!} \ ways$$