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paul c
Posts:
3
Registered:
12/6/04
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Permutations problem help
Posted:
Mar 26, 2000 10:21 PM
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dave,
this is just a matter of manipulating symbols. all you need to remember is
that P(n,r)=n!/(n-r)!.
thus,
P(n+1,3)- P(n,3) = (n+1)!/(n+1-3)! - n!/(n-3)!
= (n+1)!/(n-2)! - n!/(n-3)!
the gcd of the two fractions is, surpirse, (n-2)!. if we subtract, we get
= [(n+1)! - (n-2)n! ]/(n-2)!
we can rewrite (n+1)! as (n+1)n!, so that
= [(n+1)n! - (n-2)n! ]/(n-2)!
factoring n!,
= n! [(n+1) - (n-2)]/(n-2)!
which is
= n! [3]/(n-2)!
or simply
= 3P(n,2).
paul
p.s.
god is not so cruel as not to allow creativity. enjoy diversity in all
forms!
From: David Christman <d.m.christman@worldnet.att.net>
To: discretemath@forum.swarthmore.edu
Date: Thu, 23 Mar 2000 02:22:27 +0000
Subject: Permutations problem help
I am really stuck on the following problem:
Prove that for all integers N =>3,
P(n+1,3) - P(n,3) = 3P(n,2)
I really need some guidance about how the complete this problem. Any
help would be appreciated. Thanks.
Dave Christman
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