Possible arrangements for the word REVIEW if one E can't be

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 permutations of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutations 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 permutations 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 permutations of these letters (without restriction) would be: \(\frac{6!}{2!}=360\).

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