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.



Asymptotic expression for $j$ which satisfies $\binom{n}{j}/j! \sim k$ as $n\to\infty$
(CORRECTED EDITION)
By mucking around with expansions like Igor suggested, I found
$$ j \approx J(n,k)=
 en^{1/2} - \tfrac14\ln(n) -\tfrac12\ln(2\pi k)-\tfrac14 e^2-\tfrac12.
$$


It seems good when $k$ is small but not extremally small. For example, $\binom{100}{22}/22!\approx 6.523187$ and $J(100,6.523187)\approx 21.82764$. Similarly, $\binom{1000}{82}/82!\approx 0.1454767$ and $J(1000,0.1454767)\approx 81.93$. Probably it wouldn't be immensely difficult to give error terms.


Note that the function $\binom{n}{j}/j!$ is bell-shaped with a maximum near $j=n^{1/2}$, so actually there are two solutions for many $k$. The above approximates one of them, apparently the larger one. Finding an approximation for the smaller solution should be similar. 


ADDED: Of course the smaller solution for fixed $k$ is between $j=0$ and $j=1$ as soon as $n\gt k$, so probably it isn't interesting.


