Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.

Topic: Permutations problem help
Replies: 0

 paul c Posts: 3 Registered: 12/6/04
Permutations problem help
Posted: Mar 26, 2000 10:21 PM

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