TITLE: Asymptotic expression for $j$ which satisfies $\binom{n}{j}/j! \sim k$ as $n\to\infty$
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: This is not quite an answer to the question (Brendan McKay already gave quite a satisfactory one), but it's worth adding that 

$\binom{n}{j}/j!$ has an interesting probabilistic interpretation, namely it is the expected number of increasing subsequences of length $j$ in a (uniformly) random permutation of order $n$, and
the fact noted by the OP that this quantity goes to $0$ when $n\to\infty$ with $j>(e+\epsilon)\sqrt{n}$ can be used to give an easy proof that the expected length of a longest increasing subsequence in a random permutation of order $n$ is bounded from above by $(e+o(1))\sqrt{n}$. (And semi-conversely, the fact that $\binom{n}{j}/j!\to\infty$ for $j<(e-\epsilon)\sqrt{n}$ implies that this is the best bound that can be proved using this method. This bound is not sharp however, the correct answer for the expected length of a longest increasing subsequent is $(2+o(1))\sqrt{n}$.) See page 9 of this book.