Approximating the logarithm of the binomial coefficient We know that by using Stirling approximation:
$\log n! \approx n \log n$
So how to approximate $\log {m \choose n}$?
 A: Using $\log(n!) ≈ n \log(n)$ and the definition of the binomial coefficient,
$\log{m \choose n} ≈ m \log{m} - (m-n) \log{(m-n)} - n \log{n}$.  The same should work for any of the more precise statements of Stirling's approximation.
A: If you expand stirling's approximation for $n\choose\alpha n$, you may also notice that it can be written as 
$$\log{n\choose\alpha n} = n H(a) - \tfrac12\log(2\pi n\,\alpha(1-\alpha)) + O\left(\tfrac1n\right)$$,
where
$$H(a) = \alpha \log\tfrac1\alpha + (1 - \alpha) \log\tfrac1{1-\alpha}$$
is the binary entropy function (in nats).
If you further notice that $n\alpha(1-\alpha)$ is the variance of a binomial distribution, it becomes pretty easy to remember. (And you start to see how you might derive the central limit theorem for binomial distributions).
A: A better approximation for the logarithm of a factorial can be found by using $\log n! \approx n \log n - n$. Interestingly, the additional terms in the approximation of the binomial coefficient cancel out, and the result is the same as if you used the simpler approximation $\log n! \approx n\log n$:
$$\begin{align}
\log {n\choose m} & = \log \frac{n!}{m!(n-m)!} \\
& = \log n! - \log m! - \log (n-m)! \\
& \approx n \log n - n - m \log m + m - (n-m) \log (n-m) + n-m \\
& = n \log n - m \log m - (n - m) \log (n-m)
\end{align}$$
An even better approximation uses more terms of Stirling's approximation, giving $\log n! \approx (n+\tfrac{1}{2})\log n - n + \tfrac{1}{2}\log 2\pi$ and hence
$$\begin{align}
\log {n\choose m} &\approx (n+\tfrac{1}{2})\log n - (m+\tfrac{1}{2})\log m - (n-m+\tfrac{1}{2})\log (n-m) - \tfrac{1}{2}\log 2\pi \\
& = n\log n - m \log m - (n-m)\log(n-m) + \cdots \\
& \qquad +\tfrac{1}{2} (\log n - \log m - \log (n-m) - \log 2\pi)
\end{align}$$
where the last term is the correction from using more terms of Stirling's approximation.
