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.