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. Find the number of ways arrangement of 10 people in 7 chairs?

1. 502,600
2. 604,800
3. 705,500
4. 825,200

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

1. How many words can be formed from the letters of the word AMERICA?

1. 2,520
2. 2,846
3. 3,225
4. 3,232

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.$$= \frac{7!}{2!}$$ $$= 2,520 \ Words$$

1. How many ways of sitting arrangement can be possible for 5 Men and 4 Women, if Women want to sit together?

1. 14,265
2. 15,212
3. 16,250
4. 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, $$= 6! \times 4!$$ $$= 720 \times 24$$ $$= 17,280$$

1. How many ways of sitting arrangement can be possible for 5 Men and 4 Women, if no two Women sit together?

1. 43,200
2. 45,300
3. 47,300
4. 48,500

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, $$= 6P_4 \times 5!$$ $$= \frac{6!}{(6 - 4)!} \times 5!$$ $$= \frac{6!}{2!} \times 5!$$ $$= 43,200 \ Ways$$

1. How many four digit numbers can be formed by using 0, 1, 2, 3, 4, if repetetion is not allowed?

1. 56
2. 66
3. 86
4. 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, $$= 4 \times 4 \times 3 \times 2$$ $$= 96$$

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

1. 1,108,799
2. 1,108,800
3. 1,108,850
4. 1,108,859

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, $$= \frac{11!}{3! \times 3!} - 1$$ $$= 1,108,800 - 1$$ $$= 1,108,799 \ words$$

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

1. 280
2. 320
3. 360
4. 380

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, $$= 5! \times 3!$$ $$= 360 \ ways$$

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

1. 360
2. 320
3. 252
4. 236

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

1. 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?

1. 50
2. 60
3. 70
4. 80

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

1. In how many ways can 4 cricket players be selected from a group of 10?

1. 180
2. 199
3. 210
4. 219

Solution: 4 cricket players can be selected from the group of 10 by $$10C_4$$ ways.
Hence, $$10C_4 = \frac{10!}{(10 - 4)! \times 4!}$$ $$= \frac{10!}{6! \times 4!}$$ $$= 210 \ Ways$$