jeeteshsingh wrote:

walker wrote:

REVIEW

1) All arrangements: 6!/2 = 360 (1/2 in order to exclude double counting as [E1, E2] is the same as [E2,E1])

2) All arrangements with two E together: 5!/2 = 603) All arrangements in which one E can't be next to the other: 360 - 60 = 300.

By the way, look at this problem:

permutation-sitting-arrangement-90121.htmlit tests the same concept.

Hi Walker / Bunuel..... I do not understand why do we divide the All E arrangements by 2

We are already considering both the E as one single component... and therefore to arrange 5 different letters - R-E1E2-V-I-W... is 5!.... I guess this number of arrangement do not include both cases like R-E1E2-V-I-W & R-E2E1-V-I-W... as we haven't multiplied 5! with 2!(ways in which E1&E2 can be arranged between themselves). Hence I don't see the need to divide 5! by 2....

Can you please let me know if my reasoning is wrong?

As per the answer should be: 6!/2! (as this arrangement as 2 duplicate E) - 5! (as duplicate E is no more concern since they are one single unit for us) = 240!

THEORY:Permutations of n things of which P1 are alike of one kind, P2 are alike of second kind, P3 are alike of third kind ...................... Pr are alike of r th kind such that: P1+P2+P3+..+Pr=n is:

\(\frac{n!}{P1!*P2!*P3!*...*Pr!}\).

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is 6!/2!2!, as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be 9!/4!3!2!.

In the original question there are 6 letters out of which E appears twice. Total number of permutation of these letters (without restriction) would be: \(\frac{6!}{2!}=360\).

# of combination for which two E

are adjacent is \(5!=120\), (consider two E as one element like: {R}{EE}{V}{I}{W}: # of permutation of this 5 elements is \(5!=120\))

Total # of permutation for which two E

are not adjacent would be \(360-120=240\).

So yes, I think you are right.

Thanks Bunuel.... but I doubt Walker goes wrong on Permutation & Combinations! Walker is quite good in this topic! I hope I see the same confirmation coming from her side!