fish and bird both need an i, and rat and bird both need an r. any permutation contains only one of each letter (unless i have misunderstood the question), so A n B n C=0, A n C=0, B n C=0.
If the f in fish is in the nth position in the permutation, the number of places to put rat is (n-2)+(20-n), and there are 19! permuations of the remaining letters therefore A n B = 19! *(sum from 2 to 20 of (n-2+20-n) + sum from 23 to 26 of (n-2) + 19 =19!*( 20*18-2*18+21+22+23+24+19)=433*19!
________________________________ From: Mahesh <firstname.lastname@example.org> To: email@example.com Sent: Saturday, 19 May 2012, 7:35 Subject: Re: RE: Permutation problems
I could make some progress after your reply. I thought of doing it other way round. So now the sets are: A: Contain fish |A|=23! B: Contain rat |B|=24! C: Contain bird |C|=23! A n B: 21! A n C: 20! B n C: 21! A n B n C: 18!
A' n B' n C'= Total permutations - (A u B u C) = 26!-(23! + 24! + 23! - 21! - 20! - 21! +18!) =4.0261.... * 10 ^ 26 Let me know if this is correct. Thank you.