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I'm working on a problem of finding the maximum of $\binom{n}{r}$ in terms of $n$, where $r\le$n.

Now, I know that $\binom{n}{0}+\binom{n}{1} +...+\binom{n}{n}=2^n$, so $\binom{n}{r}$ should be no more than $2^n$. And I find that $\binom{2}{1}=2$ which is half of $2^2$ and $\binom{3}{1}=3$ which less than half of $2^3$, but it seems not a strict proving. It seems like there is a relation between $n$ and the ratio between the largest $\binom{n}{r}$ and $2^n$ with $n$ given. Maybe reciprocal? Please help or try to give some ideas how to achieve this. Thanks in advance.

MiaoYu喵宇
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    Try a few examples with $n=2,3,4,5$. The largest binomial coefficient is the middle one (or two of them) - see here, for example, or here. And the value is related to Catalan numbers, see the OEIS entry in the answer below. – Dietrich Burde Oct 01 '23 at 14:50
  • Check this post too: https://math.stackexchange.com/questions/722952/how-do-you-prove-n-choose-k-is-maximum-when-k-is-lceil-tfrac-n2-rcei – Anton Vrdoljak Oct 01 '23 at 14:58
  • For simplicity, assume that $m=2n$ is even. Then we have $$ \binom{2n}{n} \operatorname*{\sim} \frac{2^{2n}}{\sqrt{\pi n}}. $$ This is the maximum in terms of $n$, asymptotically. See this duplicate. – Dietrich Burde Oct 04 '23 at 17:50

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