Aptitude - Permutation and Combination - Discussion

7P3 * 5P2 =2520.

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

I'm not sure this is the correct forum for this question, but here goes. If I have 4 consonants and there are 5 vowel sounds that can be pronounced hard or soft, how many different pronunciations are possible?

If this is the wrong forum can you direct me to a place that could answer that question?

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.

@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.

4C2 = (3*4)/(2! * 2!).
2 VOWELS OUT OF 4.
then, why 210?

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..

Here, the answer is 302400.

Because you should 3 constants from 7 and 2 vowels from 4 and multiply then the answer is (7*6*5) * (4*3).


Later you should arrange them by multiplying with 5!.

Why can't we use permutation here instead of combination and multiplication by arrangement?

Since permutation = combination * arrangement.

But the answer is coming 2520.

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.

@Veyron999.

Best explanation.