TITLE: Solving equation involving factorial
QUESTION [7 upvotes]: I have the following problem:
$$
N^n(N-n)!=A
$$
Where $N$ and $A$ are constants.  I want to solve this equation for $n$ (for a variation of the birthday problem), but I have little experience with combinatorics.
An approximate answer is acceptable, so I've tried using Stirling's approximation, but no matter how I manipulate it, I can't get an answer to fall out.
Any guidance would be appreciated.

REPLY [5 votes]: The best method is going to depend quite a bit on the relative sizes of $N$ and $A$, but here's an idea that might work well in practice.
Divide both sides of the equation by $N!$ to yield
$$
\frac A{N!} = \frac{N^n(N-n)!}{N!} = \prod_{j=0}^{n-1} \bigg( 1 - \frac jN \bigg)^{-1}.
$$
Take logs of both sides to obtain
\begin{align*}
\log \frac A{N!} &= \sum_{j=0}^{n-1} \log \bigg( 1 - \frac jN \bigg)^{-1} \\
&= \sum_{j=0}^{n-1} \bigg( \frac jN + O\bigg( \frac{j^2}{N^2} \bigg) \bigg) \\
&= \frac{n(n-1)}{2N} + O\bigg( \frac{n^3}{N^2} \bigg).
\end{align*}
For many ranges of $N$ and $A$, this shows that $n$ will be about $\sqrt{2N\log(A/N!)}$.