Chapter 11.11, Problem 9P

The following expression occurs in statistical mechanics: $P=\frac{n!}{(np+u)!(nq-u)!}{p}^{np+u}{q}^{nq-u}$

Use Stirling’s formula to show that $\frac{1}{P}~x^{npx}{y}^{nqy}\sqrt{2\pi npqxy}$, where $x=1+\frac{u}{np},y=1-\frac{u}{nq},$ and $p+q=1.$ Hint: Show that ${(np)}^{np+u}{(nq)}^{nq-u}=n^n{p}^{np+u}{q}^{nq-u}$ and divide numerator and denominator of $P$ by this expression.