# Numbers satisfying $\binom{n}{k} = m!$

Let $k,m,n\in \mathbb{N}$ where $1 < k < n-1$. Consider the equation $$\binom{n}{k} = m!$$ which can also be equivalently written as $$n!=(n-k)!k!m!$$ The only instances I found are $\binom{4}{2} = 3!$ and $\binom{10}{3} = 5!$ I do not see any pattern coming out. As I went far out, it seemed that it is hard to find other examples as the second instance seems to be related to the problem of consecutive numbers being composed of only small primes. Is it true that these are the only instances?

Thanks.

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@Henry, thanks for the edit. – Unknown Mar 20 '11 at 18:00
There is also ${16 \choose 2}=5!$. – Henry Mar 20 '11 at 18:01
@Henry, are there more that do not include small numbers such as 2? – Unknown Mar 20 '11 at 18:05
I just wrote a script and I don't believe there are any others up to 13! (except of course $\binom{n}{0}$ and $\binom{n}{1}$). – Ben Alpert Mar 20 '11 at 18:12

You're asking about the existence of factorials in entries of Pascal's triangle. I googled for factorials pascal's triangle and found this discussion:

http://mathoverflow.net/questions/17058/factorials-in-pascals-triangle

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Thanks very much. I constructed the Pascal triangle to some extent while looking for examples but did not consider googling in those terms. – Unknown Mar 21 '11 at 14:06