Permutation and Combination

Question: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
[A].

210

[B].

1050

[C].

25200

[D].

21400

Answer: Option C

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

	= (7C3 x 4C2)
	= 7 x 6 x 5 x 4 x 3 3 x 2 x 1 2 x 1
	= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves	= 5!
	= 5 x 4 x 3 x 2 x 1
	= 120.

Required number of ways = (210 x 120) = 25200.

Video Explanation: https://youtu.be/dm-8T8Si5lg

Question:
How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

[A].

5

[B].

10

[C].

15

[D].

20

Answer: Option D

Explanation:

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

Required number of numbers = (1 x 5 x 4) = 20.

Question:
A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

[A].

32

[B].

48

[C].

64

[D].

96

Answer: Option C

Explanation:

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

Required number of ways	= (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
	= 3 x 6 x 5 + 3 x 2 x 6 + 1 2 x 1 2 x 1
	= (45 + 18 + 1)
	= 64.

Question: In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
[A].

159

[B].

194

[C].

205

[D].

209

Answer: Option D

Explanation:

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required number of ways	= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
	= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
	= (6 x 4) + 6 x 5 x 4 x 3 + 6 x 5 x 4 x 4 + 6 x 5 2 x 1 2 x 1 3 x 2 x 1 2 x 1
	= (24 + 90 + 80 + 15)
	= 209.

Question:
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

[A].

564

[B].

645

[C].

735

[D].

756

Answer: Option D

Explanation:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

Required number of ways	= (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
	= 7 x 6 x 5 x 6 x 5 + (7C3 x 6C1) + (7C2) 3 x 2 x 1 2 x 1
	= 525 + 7 x 6 x 5 x 6 + 7 x 6 3 x 2 x 1 2 x 1
	= (525 + 210 + 21)
	= 756.

Question:
In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

[A].

266

[B].

5040

[C].

11760

[D].

86400

Answer: Option C

Explanation:

Required number of ways	= (8C5 x 10C6)
	= (8C3 x 10C4)
	= 8 x 7 x 6 x 10 x 9 x 8 x 7 3 x 2 x 1 4 x 3 x 2 x 1
	= 11760.

Question: In how many different ways can the letters of the word ‘DETAIL’ be arranged in such a way that the vowels occupy only the odd positions?
[A].

32

[B].

48

[C].

36

[D].

60

Answer: Option C

Explanation:

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under:

(1) (2) (3) (4) (5) (6)

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.

Number of ways of arranging the vowels = 3P3 = 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = 3P3 = 3! = 6.

Total number of ways = (6 x 6) = 36.

Question:
In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged so that the vowels always come together?

[A].

10080

[B].

4989600

[C].

120960

[D].

None of these

Answer: Option C

Explanation:

In the word ‘MATHEMATICS’, we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

Number of ways of arranging these letters =	8!	= 10080.
	(2!)(2!)

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters =	4!	= 12.
	2!

Required number of words = (10080 x 12) = 120960.

Question: In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?
[A].

810

[B].

1440

[C].

2880

[D].

50400

Answer: Option D

Explanation:

In the word ‘CORPORATION’, we treat the vowels OOAIO as one letter.

Thus, we have CRPRTN (OOAIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

Number of ways arranging these letters =	7!	= 2520.
	2!

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged

in	5!	= 20 ways.
	3!

Required number of ways = (2520 x 20) = 50400.

Video Explanation: https://youtu.be/o3fwMoB0duw

Question: In how many different ways can the letters of the word ‘LEADING’ be arranged in such a way that the vowels always come together?
[A].

360

[B].

480

[C].

720

[D].

5040

Answer: Option C

Explanation:

The word ‘LEADING’ has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

Video Explanation: https://youtu.be/WCEF3iW3H2c