Aptitude - Permutation and Combination - Discussion

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?

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.

How can we find permutation and combination problem ?

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..

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.

NO. It's not because of the letters (and not person) we should multiply with 5!, but it's about what the question said; how many words can be FORMED? that's we call PERMUTATION. At first we might have thought it's just COMBINATION because we didn't notice the last phrase of the question said how many words can be FORMED. So it's considered as both PERMUTATION and COMBINATION.

I don't know why all are confusing as permutation and combination.

I think this question"how many words can be FORMED " is clearly combination.

When I saw this question I was like it will be permutation since arrangement of words is important and I did 7P3*4P2(just like other sums).

Why does the method differ here !

@Harsha.

Here the asking question is group of words (use combination) not to asking about the order of words so don't use the permutation. OK.

Here we should use permutations we need to use combinations and just selecting the consonants and vowels from the given 7 consonants and 4 vowels, its not necessary to multiply with 120 there so according to me, its 210 just selecting those consonants and vowels from the given one not more than that.