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