Approximating $\frac{(kn)!}{(n!)^k}$

Question

Is there any approximations for the form $$\frac{(kn)!}{(n!)^k},$$ where $n$ and $k$ are positive integers? $n$ is not necessary much larger than $k$?

Answer 1

The usual form of Stirling's Approximation is

$$m! \approx \sqrt{2 \pi m} \left(\frac{m}{e}\right)^m,$$

which is a very good approximation for even modest $m$. Substituting using this approximation gives

$$\frac{(kn)!}{n!^k} \approx \frac{\sqrt{2 \pi (kn)} \left(\frac{kn}{e}\right)^{kn}}{\left[\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n\right]^k} = (2 \pi n)^{\frac{1 - k}{2}} k^{kn + \frac{1}{2}}.$$ This approximation should be quite good again for $n, k$ that are not too small, and one could produce rather tight error estimates using the estimates given in the linked article.