# Demonstration of a divisibility rule

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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That's not what your formula says, since there is a separate factor $(m-n)!$ for every choice of $i$. For instance for $n=1$ and $m=3$ your product is $\frac1{2!}\times\frac2{2!}\times\frac3{2!}=\frac68=\frac34$, which is not an integer. I'll now make the formula match your words. Oops, JavaMan already did that. – Marc van Leeuwen Feb 14 '12 at 14:20
@Marc: I incorrectly edited the post. That mistake belongs to me. I have since fixed the post. – JavaMan Feb 14 '12 at 14:22

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So your response to his/her challenge was to ask someone else to do it? – Cam McLeman Feb 14 '12 at 14:59
I looked at some divisibility rules on Wikipedia and I got demotivated! – Tom Dwan Feb 14 '12 at 15:37

## 1 Answer

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.

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