Bunuel wrote:
anilnandyala wrote:
We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: .
Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be
CAN ANYONE EXPLAIN LAST STEP
THANKS IN ADVANCE
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.
I didn't understand the statement "Hence exactly in half of these cases D will be to the right of C and in half of these cases to the left." Asked myself Why? How? (similar situation with the arrangement of Frankie and Joe - question)
The following helped me understand: The letters A, B and C can be arranged in 3 slots in 3*2*1 ways.
ABC, ACB, BAC, BCA, CAB and CBA. Under normal circumstances C comes before B thrice; B comes before C thrice. For a specific case we consider whatever is required.
I hope my understanding is right. If yes, I hope it helps someone.
-Arvind