TITLE: solve for variable in combination
QUESTION [5 upvotes]: i have the combination ${n\choose 11}=12376$ and am looking to solve for $n$.  it turns out to be $17$.  of course can use brute force approach where just plug numbers in for $n$ but am looking for a cleaner method?
so ${{n!}\over {11!(n-11)!}}=12376$
$n(n-1)(n-2)...(n-10)=11!(12376)$
by the pigeonhole principle can eliminate $10!$ and be left with
$n(something)=11(12376)=11(2^3)(7)(13)(17)$
thanks.

REPLY [2 votes]: As an approximation, I would note that the AM-GM gives
$$
\begin{align}
12376
&=\frac{n(n-1)\cdots(n-9)(n-10)}{11!}\\
&\lesssim\frac{(n-5)^{11}}{11!}
\end{align}
$$
Therefore,
$$
n\gtrsim5+(11!\cdot12376)^{1/11}=16.5629
$$
Trying $n=17$ gives $\binom{17}{11}=12376$