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.