Aptitude - Permutation and Combination - Discussion

Finally multiplied with 5!, because, each word is consisted of 5 distinct letters. Had it been persons in place on letters, then 5! multiplication would not have been applied.

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..

How can we find permutation and combination problem ?

Hi,

I always find problem in using C and P in formula I could not distinguish when to use nCr and when to use nPr please help in finding difference.

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?

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).

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.

Why we need multiply 120 with 210 is it necessary?

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.

Miss suchi, from given question it was clear that out of 7 consonants 3 consonants and 2 vowels of 4 vowels could considered.