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Aptitude - Permutation and Combination

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Exercise : Permutation and Combination - General Questions
1.
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?
564
645
735
756
None of these
Answer: Option
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.

2.
In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
360
480
720
5040
None of these
Answer: Option
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

3.
In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?
810
1440
2880
50400
5760
Answer: Option
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

4.
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
210
1050
25200
21400
None of these
Answer: Option
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

5.
In how many ways can the letters of the word 'LEADER' be arranged?
72
144
360
720
None of these
Answer: Option
Explanation:

The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.

Required number of ways = 6! = 360.
(1!)(2!)(1!)(1!)(1!)

Video Explanation: https://youtu.be/2_2QukHfkYA

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