Crossposted from https://math.stackexchange.com/questions/2605895/asymptotics-for-partial-sum-of-product-of-binomial-coefficients

For some $ c<n$ and $ 2c\leq x\leq 2n$ , are there references or previous results for determining the asymptotics (as $ n\to\infty$ ) of the partial sum $ $ \sum_{k=x-c}^c\binom{n}{k}\binom{n}{x-k} $ $ or equivalently if $ c=n\lambda_1$ and $ x=2n\lambda_2$ , for constants $ 0<\lambda_2\leq\lambda_1<1$ $ $ \sum_{k=2n\lambda_2-n\lambda_1}^{n\lambda_1}\binom{n}{k}\binom{n}{2n\lambda_2-k} $ $

I don’t think I can just apply Stirling’s approximations to the binomial coefficients individual and take the sum and product.

**EDIT**

Could someone comment if this is a valid attempt?

Using @robjohn’s solution in this post, let $ $ a_k=\binom{n}{k}\binom{n}{2n\lambda_2-k} $ $ Then letting $ k=n\lambda_2+j$ , $ $ \log\left(\frac{a_{k+1}}{a_k}\right)=-\frac{2j}{n\lambda_2(1-\lambda_2)}+O(n^{-1}) $ $ Thus, $ $ a_k=a_{n\lambda_2}\exp\left(-\frac{2j^2}{n\lambda_2(1-\lambda_2)}+O(j/n)\right) $ $ Estimating $ $ a_{n\lambda_2}\sim C(\lambda_2)=\frac{1}{2\pi n\lambda_2(1-\lambda_2)}(1-\lambda_2)^{-2n}\left(\frac{1-\lambda_2}{\lambda_2}\right)^{2n\lambda_2} $ $ by Stirling’s formula and using Riemann integral for the exponential, $ $ \sum_{j=-n(\lambda_1-\lambda_2)}^{n(\lambda_1-\lambda_2)}\exp\left(-\frac{2j^2}{n\lambda_2(1-\lambda_2)}+O(j/n)\right)=\sqrt{n\lambda_2(1-\lambda_2)}\int_{-\infty}^{\infty}\exp\left(-2t^2\right)dt(1+O(1/n)) $ $ we have \begin{eqnarray} \sum_{k=2n\lambda_2-n\lambda_1}^{n\lambda_1}\binom{n}{k}\binom{n}{2n\lambda_2-k}&\sim& C(\lambda_2)\sqrt{n\lambda_2(1-\lambda_2)}\sqrt{\pi/2}\ &=&\frac{1}{2\sqrt{2\pi n\lambda_2(1-\lambda_2)}}(1-\lambda_2)^{-2n}\left(\frac{1-\lambda_2}{\lambda_2}\right)^{2n\lambda_2} \end{eqnarray} Substituting back $ c=n\lambda_1$ and $ x=2n\lambda_2$ , and noticing Stirling’s formula for $ \binom{2n}{x}$ , we get $ $ \sum_{k=x-c}^c\binom{n}{k}\binom{n}{x-k}\sim\frac{1}{\sqrt{2}}\sqrt{\frac{2n}{2\pi x(2n-x)}}\left(\frac{2n}{2n-x}\right)^{2n}\left(\frac{2n-x}{x}\right)^x\sim \frac{1}{\sqrt{2}}\binom{2n}{x} $ $ To me this is very interesting that it doesn’t involve $ c$ , which disappeared when estimating with the Riemann integral above. However, after plugging in a couple of values in Mathematica, the approximation on the right hand side doesn’t always give an accurate approximation to the partial sum.