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The mathematical form for nCr is (n!)/(r!(n-r)!)

How does this form ensure that nCr is indeed a whole number. Is there a mathematical proof?

Hashken
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    Have a look at Gone's answer to http://math.stackexchange.com/questions/2158/division-of-factorials/2192#2192 – Arthur Nov 05 '13 at 06:03

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The nice thing is that it is the combinatorial argument that $\frac{n!}{r!(n-r)!}$ counts something (i.e., the number of ways to choose $r$ items from $n$ unordered items) that proves (in the most rigorous sense) that it is a natural number. What are proof is needed?

Ittay Weiss
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  • The argument that a combinatorial argument should be a natural number as it is basically counting something, is perfectly valid. But, I was looking for a more rigorous mathematical proof that the given expression always gives a natural number, not just a logical argument. – Hashken Nov 06 '13 at 06:04
  • do you consider a logical argument to be lacking in mathematical rigor??? – Ittay Weiss Nov 06 '13 at 07:32