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A friend of mine who's studying mathematics challenged me to demonstrate that:

For given integer numbers $n$ and $m$, we can say

$\left(\prod_{i=n}^m i\right)/{(m-n)!} =Z,$

where $Z$ is some integer. In other words, the product of $n(n+1)(n+2)...m$ can be divided by the factorial of the difference.

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    I looked at some divisibility rules on Wikipedia and I got demotivated!2012-02-14

1 Answers 1

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Since the quantity

$ {m \choose n} = \frac{m!}{n!(m-n)!} = \frac{m(m-1)\dots (n+2)(n+1)}{(m-n)!} $

is always an integer, then it follows that $n$ times that quantity is also an integer.