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Old 11-28-2007, 08:09 PM
pzhon pzhon is offline
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Join Date: Mar 2004
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Default Hyperfactorial Problem

We're used to dealing with n!
How about n?

Define n? = 1! 2! ... (n-1)!
Do not include an n! factor.
So, 1? = 1, 2? = 1, 3? = 2, 4? = 12, and 5? = 288.

It is known that for any positive integers a, b, and c, that f(a,b,c)=

(a+b+c)? a? b? c?

counts something. Prove that f(a,b,c) is an integer without using that it counts something. (Note that for a=1, f(a,b,c) specializes to b+c choose b,c.)

Please post solutions in white.
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