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Q&A #3398

Teachers' Lounge Discussion: Permutations and counting arrangements

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From: Loyd <loydlin@aol.com>
To: Teacher2Teacher Public Discussion
Date: 2004050712:26:07
Subject: Re: Re: Re: HELP;removing common factor!



Sarah, that is a nice proof.  At first I was confused but finally saw
the light.  On your third step you factored out the common factor.

On my proof, I noticed that a!/(a-1!) = a.  That may not be obvious to
some, but here is an example.  

Let a=7, then a!=1x2x3x4x5x6x7
Then (a-1)!= 1x2x3x4x5x6

So that 1x2x3x4x5x6x7
        -------------- = 7
        1x2x3x4x5x6




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