Approximation of binomial coefficient using Stirlings formula

Question

We have the approximation: $\ln x! \simeq x \ln x - x + \tfrac{1}{2} \ln (2 \pi x)$.

I want to estimate $\ln \binom{N}{r} = \ln\left(\frac{n!}{(N - r)! r!}\right) = \ln(n!) - \ln((N - r)!) - \ln(r!)$. The result I want to get is $$(N - r) \ln(N / (N - r)) + r \ln(N / r).$$

but I don't get that result, I have got $$N \ln N - r \ln r - (N - r) \ln (N - r) + \tfrac{1}{2} \ln (2 \pi N) - \tfrac{1}{2} \ln (2 \pi r) - \tfrac{1}{2} \ln (2 \pi (N - r))$$

How can I proceed from here to get the desired result?

Thanks.

Answer 1

You have:

$\begin{align*} \ln \binom{N}{r} &\sim (N - r) \ln N - (N - r) \ln (N - r) - r \ln r + r \ln N + \frac{1}{2} \ln (2 \pi N) - \frac{1}{2} \ln (2 \pi r) - \frac{1}{2} \ln (2 \pi (N -r)) \\ &= (N - r) \ln \frac{N}{N - r} + r \ln \frac{N}{r} + O(\ln N) \end{align*}$

Unless $r$ is very near $0$ or $N$, you can disregard the lower-order terms.