Simplifying factorial with different coefficient

Question

$$\frac{(pn)!}{(qn)!},\quad p\not = q$$

If possible, how could I simplify the above factorial?

Answer 1

If you need an exact calculation, I suggest, just as @Reety commented, to write $$A=\frac{(pn)!}{(qn)!}=\frac{\Gamma( pn + 1)}{\Gamma( qn + 1)}\implies \log(A)=\log (\Gamma (n p+1))-\log (\Gamma (n q+1))$$ and use the $\text{log$\Gamma $}(x)$ function which is available almost in any computing environment [Excel has it; it is GAMMALN( x )]. An then, for sure $A=e^{\log(A)}$.

If you want an approximate value, as Ross Millikan answered, Stirling approximation is the most convenient $$\log(A)\approx(p-q) n (\log (n)-1) + (p \log (p)-q \log (q))n+\frac{1}{2} \log \left(\frac{p}{q}\right)+\frac{q-p}{12 p q}\frac 1n+O\left(\frac{1}{n^3}\right)$$ In these forms, $p$ and $q$ do not require to be integer numbers.

Trying with $n=100$, $p=\pi$, $q=e$, the approximation would give $A=2.75535890270\times 10^{104}$ while the exact value is $2.75535890283\times 10^{104}$

Answer 2

Depending on your needs, Stirling's approximation may be of help. Here it gives $$\frac {(pn)!}{(qn)!}\approx \frac{(pn)^{pn}}{(qn)^{qn}}e^{(q-p)n}\sqrt{\frac pq}$$