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Re: RE: Permutation problems
Posted:
May 19, 2012 12:03 PM
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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!
A'nB'nC'=26!-23!-24!-23!+433*19!
________________________________
From: Mahesh <discussions@mathforum.org>
To: discretemath@mathforum.org
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.
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