# Asymptotic for binomial coefficient with square root

I'm looking for asymptotic estimate for the binomial coefficient: $$\ln{\binom{n}{[\sqrt{n}]}}$$ I assume Stirling's approximation can help, but I'm not sure I will get any good estimation with this approach. Is there any good way to make an estimation for this coefficient? Thanks in advance.

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Use $\ln(n!)=(n+1/2)\ln(n)-n+O(1)$ for all the three terms. – Shitikanth Oct 5 '12 at 20:14

Using Shitikanth's hint I think you're going to be coming up with $$\text{ln}{n \choose [\sqrt{n}]}\approx\text{ln}\left(\frac{n^{n+\sqrt{n}/2+3/4}}{\left(n-\sqrt{n}\right)^{1/2-\sqrt{n}+n}}\right).$$
@Kos Well, I started with $\text{ln}{n \choose [\sqrt{n}]}=\text{ln}{(n)!}-\text{ln}(n - [\sqrt{n}])! + \text{ln}[\sqrt{n}]!$ and then use Shitikanth's expansion for all three terms, then I used the fact that $\sqrt{n}=n^{1/2}$ to help simplify at the end. – Alexander Gruber Oct 6 '12 at 4:39
It may just be that ${n \choose [\sqrt{n}]}$ is an important term in a lot of algorithms and that's why they put it in there. For what it's worth my answer looks right based on the plot of it: i.imgur.com/JOPFU.gif – Alexander Gruber Oct 6 '12 at 4:41