Suppose $k>0$ is some fixed constant, and $n$ is a positive integer tending to infinity. Find $j\equiv j(n,k)$ such that $$ \frac{\binom{n}{j}}{j!} \sim k. $$

The asymptotic expression for $(n!)^{-1}$ (http://stackoverflow.com/questions/3084937/how-to-calculate-the-inverse-factorial-of-a-real-number) was initially helpful, but upon rearranging we have equivalently to solve for $j$ in $$ k\ j!^2 = n(n-1)\cdots(n-j+1),$$ and in this expression both sides depend on $j$.

I've tried Stirling's approximation, and assuming I did the calculations correct, $e \sqrt{n}$ is interesting since $j \ge e \sqrt{n}$ implies $\binom{n}{j}/j! \to 0$ whereas $j < e \sqrt{n}$ implies $\binom{n}{j}/j! \to \infty$.

In de Bruijn's book there are implicit methods, but I wasn't able to successfully apply any of the approaches.

Suppose $k>0$ is some fixed constant, and $n$ is a positive integer tending to infinity. Find $j\equiv j(n,k)$ such that $$ \frac{\binom{n}{j}}{j!} \sim k. $$

The asymptotic expression for $(n!)^{-1}$ (https://stackoverflow.com/questions/3084937/how-to-calculate-the-inverse-factorial-of-a-real-number) was initially helpful, but upon rearranging we have equivalently to solve for $j$ in $$ k\ j!^2 = n(n-1)\cdots(n-j+1),$$ and in this expression both sides depend on $j$.

I've tried Stirling's approximation, and assuming I did the calculations correct, $e \sqrt{n}$ is interesting since $j \ge e \sqrt{n}$ implies $\binom{n}{j}/j! \to 0$ whereas $j < e \sqrt{n}$ implies $\binom{n}{j}/j! \to \infty$.

In de Bruijn's book there are implicit methods, but I wasn't able to successfully apply any of the approaches.