Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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
up vote 7 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.