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 4 of 7.

Sachin Jain said:   1 decade ago
Dear friends

Keep in mind first that it is a permutation problem becoz the order of letters matter in order to make different words. Secondly out of 7 consonants we have to select 3 so when we select a consonant out of 7 it will not again be considered to be selected so the repetition is not allowed. Now the formula for Permutation without repetition is (n!)/(n-r)!. So it will be 7P3*4P2.

=(7!/4!)*(4!/2!)=2520 (not 25200).

Mark Burnham said:   8 years ago
This is a question of permutation, as the order in which letters are arranged changes the final result.

Therefore = 7P3*4P2 = 2520.

However, there are 10 different ways that the consonants and vowels can be arranged.

So, the answer is 2520*10,
=2520.

Navin said:   1 decade ago
Hey they anly asked us how many ways to form the letters they didint ask how many ways to arrange it the answer is wrong it should be 210.

Student said:   1 decade ago
Why we need multiply 120 with 210 is it necessary?

M.V.KRISHNA/PALVONCHA said:   1 decade ago
Hi suchi.

Given 3c out of 7c

2v out of 4v.

So, required number of letters are 5. they can be arranged among

themselves in 5!.

Hope you understood.

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.

Suchi said:   1 decade ago
c = consonants
v = vowels

Why has the num of letters has to be 5?

Can't it be 4c+3v(7 letter) or 5c+4v(9 letters)

Then how can v take 5! ?

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

Student said:   1 decade ago
Can't we apply permutation directly here since in this question arrangement of words is to be done. i.e.

7P3*4P2 = 2520

One zero less to match the answer of 25200 :(

Anil said:   2 decades ago
Why multiply with 5?


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