Aptitude - Permutation and Combination - Discussion

Discussion Forum : Permutation and Combination - General Questions (Q.No. 4)
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

Discussion:
65 comments Page 5 of 7.

Rishi said:   10 years ago
In question instead of how many words there should be how many arrangements.

Otherwise, a solution is wrong according to question.

Gezoi said:   8 years ago
7P3 * 5P2 =2520.

Then considering combining them accounting for repetitions we get = 5!/(3!*2!)=10.
Therefore =2520*10=25200.
(1)

Machender said:   1 decade ago
Miss suchi, from given question it was clear that out of 7 consonants 3 consonants and 2 vowels of 4 vowels could considered.

Hehe said:   8 years ago
I don't think that it's supposed to be multiplied by 5! since it doesn't ask for the arrangement of the letters.

Atul Moundekar said:   9 years ago
This question is only based on a combination, they never told to find the arrangements. So the answer is 210.

Achu said:   1 decade ago
What's the need to take 5 here? They didn't mention any groupings here. Then what's the purpose to group?

Pranay said:   1 decade ago
The question doesn't mention "distinct" consonants or vowels. So repetition should be allowed.

Kaustav said:   1 decade ago
Combination 3 consonant and 2 vowels is still left. By adding forms 5! and then rest to find the total.

May said:   8 years ago
Why do 5!?

What if the consonants or vowels have repetition? Does it necessary to multiply with 5!?
(5)

Anomiee said:   1 year ago
If the question is how many words we can form then why need to find the number of ways?
(6)


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