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 2 of 7.

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)

Aadesh said:   8 years ago
I agree @Ankita.

That as we are picking up the words by applying combination means that we are forming the words therefore answer. It Should be 210.
(1)

Ankita said:   8 years ago
@Himatheja.

We have to make word from both consonants and vowel that's why we have to multiply combination of consonants with vowel also,
7c3*4c2.
= 35*6.
= 210.
(1)

Himatheja said:   8 years ago
4C2 = (3*4)/(2! * 2!).
2 VOWELS OUT OF 4.
then, why 210?
(1)

Veyron999 said:   1 decade ago
This question can be solved by two methods
1) using combinations:

we have to select 3 consonants from 7 and 2 vowels from 4,
we will use 7C3 x 4C2, this implies that for eg. consonants are B C D F G H J, then
if BCD is there then CBD and DBC won't be there and so on..
but they will make different words, so
we arrange them later on by multiplying by 5! ..
eg. one combination will be BCD AE this can be arranged in 5! ways and so on...
so ans. is 7C3 x 4C2 x 5! = 25200

2) using permutations:

unlike previous method if we use 7P3 we also consider BCD, CBD and all possible arrangements...

then we get combinations of the type CCCVV (C- consonant, v- vowel)
but they can be arranged in 5! ways eg- CVCVC, VVCCC, etc..
but using permutation, we have already accounted for those combinations in which all 3 consonants occur together and 2 vowels occur together.. so to eliminate those cases, we need to divide it by the repetitions i.e. divide by 3!2! (for consonants and 2 for vowels) and then multiply this with the initial result..
so ans is 7P3 x 4P2 x 5!/(2!3!) = 25200

tried to explain as clearly as possible..hope i was of some help..
(1)

Litu said:   1 decade ago
Hi,

I do agree with Sachin Jain as it is a permutation query. We can arrange it in any way using using 3 consonants out of 7 & 2 vowels out of 4.

Ex: If there is bcdfghj then we can do it bcd n also cbd so it is permutation type not combination so the answer should be 7P3 * 4P2 = 2520

But again we have to arrange these 2 groups together. So I'm not getting form this point.

Can anyone clear the same?

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.

Mahima said:   9 years ago
Why can't we use permutation here instead of combination and multiplication by arrangement?

Since permutation = combination * arrangement.

But the answer is coming 2520.

Nitish said:   9 years ago
As the consonants and vowels are not specified and we only have to find how many ways we can arrange them i.e. 210 the Order does not matter. If the letters were specified then only we can apply permutation here.

For example, if the question was 7 men and 4 women, we wouldn't multiply with 5!.
Hence, the correct answer would be 210.

Sundeep said:   9 years ago
@Veyron999.

Best explanation.


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