Asymptotics of partial sum of binomial coefficients

For some fixed $ 0<p<1$ , with $ q=1-p$ , let $ np\leq c\leq np+\sqrt{2npq-2n\log\log n}$ and $ 2np\leq x\leq 2np+2\sqrt{2npq-2n\log\log 2n}$ . I’m trying to get asymptotics 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 $ p\leq\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} $ $

My initial attempt was to adapt @robjohn’s solution in this post.

First we focus on $ $ 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) $ $ By Stirling’s approximation, we have $ $ a_{n\lambda_2}\sim \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}=C(\lambda_2) $ $ Then using the 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} $ $

However, I’m not entirely convinced that this asymptotic holds for the ranges given for $ c$ and $ x$ . Is there a refinement for this result?