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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?

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  • $\begingroup$ What would you call a pair of numbers $n,m$ s.t. $n\choose m$ is a factorial? $\endgroup$
    – Gregory Grant
    Jun 11 '15 at 21:00
  • $\begingroup$ @GregoryGrant I'm not sure. Is there already a name for such things? (Read: is there an answer out there and you're leading me to it?) $\endgroup$
    – Cameron Williams
    Jun 11 '15 at 21:01
  • $\begingroup$ I recall there was a question on MSE about writing factorials as products of other factorials (which this is a special case of). I don't remember if anything useful enough came of it to answer this question. $\endgroup$
    – anon
    Jun 11 '15 at 21:04
  • $\begingroup$ @anon Hmm that's a shame. I've since computed up to $n=100$ and $m=500$ and no others have appeared. $\endgroup$
    – Cameron Williams
    Jun 11 '15 at 21:08
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    $\begingroup$ These have to be fairly rare. Consider: the largest binomial coefficient in row $m$, the central coefficient, is smaller than $2^m$ (since the sum of all the coefficients in that row is of that size). So for a given value $n$, since $n!\approx (n/e)^n$ (I'm neglecting any subexponential factors here), the minimal value of the row $m$ where we can expect to find $n!$ satisfies $2^m\approx (n/e)^n$ or $m\approx n\log_2n$. But then we have to 'dodge' all of the primes between $n$ and $m\approx n\log_2n$ as potential factors for our binomial coefficients, and this is going to be very hard to do. $\endgroup$
    – Steven Stadnicki
    Jun 11 '15 at 21:12
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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.

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  • $\begingroup$ I was too busy searching for "binomial coefficients" to even consider the more general problem so I missed that post entirely. Tunnel vision got the better of me on this one. This is really neat. $\endgroup$
    – Cameron Williams
    Jun 11 '15 at 21:26
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    $\begingroup$ I do wonder whether the additional structure here offers enough to conclude that there are no more solutions for this special case. It seems unlikely (since in particular you would presumably need some results about e.g. $n$ and $n+1$ not being simultaneously smooth that could be really hard) but this case does seem more within reach than the general question of products-of-factorials. $\endgroup$
    – Steven Stadnicki
    Jun 11 '15 at 21:40
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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.

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