When is a binomial coefficient a factorial, i.e. when is $\binom{m}{j} = n!$ for some $n,m,j$? As stated in the title: when is a binomial coefficient a factorial, i.e. when is $\binom{m}{j} = n!$ for some $m,j,n$? I was thinking about this problem a couple of days ago because in all my years of being a mathematician (amateur or otherwise), I only noticed a handful of these.
Of course there are a couple of trivial cases to negate: the case when $n$ arbitrary $m = n!$ and $j=1$, likewise when $m$ arbitrary and $j=0$ and $n=0,1$.
I decided to write a Mathematica code to check this for me which I can make available for anyone who is interested. I decided to compute up to $30!$, i.e. $n=30$, and $m$ up to $110$. There were only $3$ binomial coefficients that were factorials (that were not trivial) up to binomial coefficient symmetry:


*

*$\dbinom{4}{2} = 6 = 3!$

*$\dbinom{10}{3} = 120 = 5!$

*$\dbinom{16}{2} = 120 = 5!$


Given how early these occur in the computations, it suggests that there are only finitely many such binomial coefficients - neglecting the trivial cases. Is there anything known in this direction? Are these actually the only ones or are there more lurking out there?
 A: Here’s the abstract of Florian Luca, On factorials which are products of factorials, Mathematical Proceedings of the Cambridge Philosophical Society, Volume 143, Issue 03, November 2007, pp 533-542:

In this paper we look at the Diophantine equation
$$n!=\prod_{i=1}^ta_i!\qquad n>a_1\ge a_2\ge\cdots\ge a_1\ge 2\;.$$
Under the $ABC$ conjecture, we show that it has only finitely many nontrivial solutions. Unconditionally, we show that the set of $n$ for which the above equation admits an integer solution $a_1,\ldots,a_t$ is of asymptotic density zero.

See also this question, especially the accepted answer by Gerry Myerson, which indicates that you’ve found the only examples known as of 2004.
A: I think they are quite rare.
We have
$$
\binom{m}{j}=\frac{m!}{j!(m-j)!}
$$
If this has to be equal to $n!$ for some $n$ the denominator has to equal $m^\underline{m-n}=m\cdot (m-1)\cdots (m-(m-n)+1)=m\cdot (m-1)\cdots (n+1)$.
Typically there will be some primes in that product, of a size comparable to $m$, they will only occur in the denominator if $j$ is a big number or if $m-j$ is big making $j$ small.
