gmatJP wrote:
why divide by only 2!3! ? there are 8spaces(letters) so dont you divide by 8!?
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 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \(\frac{8!}{2!3!}=3360\).
LM wrote:
Can you please explain the logic or how you could deduce quickly the following:-
Now, in half of these cases D will be to the right of C and in half of these cases to the left
Obviously C and D can have ONLY TWO positions: C to the right of C OR to the left, how else?
Now, why should C (or D) be in more cases to the right (or to the left) of D (C)? Does probability favors either of these letters? No. Hence exactly in half of these cases D will be to the right of C and in half of these cases to the left.
Hope it's clear.
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