Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Use $\ln(n!)=(n+1/2)\ln(n)-n+O(1)$ for all the three terms. – Shitikanth Oct 5 '12 at 20:14
up vote 1 down vote accepted

Using Shitikanth's hint I think you're going to be coming up with $$\text{ln}{n \choose [\sqrt{n}]}\approx\text{ln}\left(\frac{n^{n+\sqrt{n}/2+3/4}}{\left(n-\sqrt{n}\right)^{1/2-\sqrt{n}+n}}\right).$$

share|cite|improve this answer
Thank you. But I'm thinking if there is any other hints for this? I'm confused with sqrt(n), because if there is no special hint, why didn't they use any other parameter or number (like a, or whatever) instead of sqrt? – Kos Oct 6 '12 at 4:27
@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: – Alexander Gruber Oct 6 '12 at 4:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.