TITLE: Factorials in Pascals Triangle
QUESTION [36 upvotes]: Hi,
I asked this question of Keith Conrad, and he suggested that I try posting here.  One of my students observed that the only instances of factorials in the interior of Pascal's triangle are $\binom{4}{2}=3!$ and $\binom{10}{3}=\binom{16}{2}=5!$.  I checked the first 500 rows, and he's right up to that point.
This is a special case of the apparently unsolved problem of finding non-trivial solutions to n!=a!b!c!...  The special feature here is that I need (a+b)!=a!b!c!, and that seems like a special enough case to have been treated by someone.  Unfortunately, literature searches have been fruitless, because every paper about Pascal's triangle contains the word "factorial" somewhere.
My best idea (which I can't make work) is to show that the powers of 7 in the equation (a+b)!=a!b!c! can't be made to match up unless neither side is a multiple of 7.  Then exhaustive searching can show that the above are the only non-trivial solutions.  
Thanks greatly for any ideas that anyone might have.

REPLY [8 votes]: The latest issue of the Journal of Number Theory (February 2016 issue) contains a paper by Nair and Shorey which gives a conditional resolution of this problem.  They consider the problem of finding solutions to 
$$ 
n! = a_1! a_2! \cdots a_t!,
$$ 
with $n>a_1 \ge a_2 \ge \ldots a_t>1$ and $t>1$.   Obviously we can assume that $a_1 \le (n-2)$, for if $a_1 =n-1$ we are just looking at $n=a_2!\cdots a_t!$.  
Then Theorem 4 of the cited paper shows that assuming Baker's explicit $abc$-conjecture, the only solutions to this equation are
$$ 
7!3!3!2!=9!; 7!6!=10!; 14!5!2!=16!.
$$
Here the explicit $abc$ conjecture of Baker states that if $a$, $b$ and $c$ are coprime positive integers with $a+b=c$ and $\omega$ denotes the number of distinct prime factors of $abc$ then 
$$
c< \frac{6}{5} N \frac{(\log N)^{\omega}}{\omega!},
$$ 
with 
$$ 
N = \prod_{p|abc} p 
$$ 
denoting the radical of $abc$.