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We know that by using Stirling approximation: $\log n! \approx n \log n$

So how to approximate $\log {m \choose n}$?

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2 Answers 2

up vote 13 down vote accepted

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

$$\begin{align} \log {n\choose m} & \approx (n+\tfrac{1}{2})\log n - (m+\tfrac{1}{2})\log m - (n-m+\tfrac{1}{2})\log (n-m) - \tfrac{1}{2}\log 2\pi \end{align}$$

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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.

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