Showing $\sum_{\alpha=0}^k \binom{k+15}{\alpha + 10}\binom{k}{\alpha}=\frac{2^{2k+15}(k+6)(k+7)\left(\frac{2k+15}{2}\right)!}{\sqrtπ (k+10)!}$ I want to simplify $\sum_{\alpha=0}^k \binom{k+15}{\alpha + 10}\binom{k}{\alpha}$ but this can't be directly simplified using Vandermonde Chu Identity. Wolfram shows a nice closed form expression but I can't get to it.
My attempt
$\sum_{k=0}^n \binom{k+15}{5+k-\alpha}\binom{k}{\alpha}=\binom{2k+15}{5+k}$ by thinking of this sum as choosing a set of 5+k objects from 2k+15 objects in total but I feel that this isn't correct as some terms are missing


Prove that $$\binom{2k+15}{k+5}=\frac{2^{2k+15}(k+6)(k+7)\left(\frac{2k+15}{2}\right)!}{\sqrtπ (k+10)!}$$

Number of $2$'s in $\left(\frac{2k+15}{2}\right)!=\frac{2k+15+1}{2}=k+8$. So, the RHS becomes
\begin{align*}
&\Rightarrow\frac{2^{k+7}(k+6)(k+7)(2k+15)!!\ \Gamma\left(\frac12\right)}{\sqrtπ(k+10)!}\\
&=\frac{2^{k+5}(k+5)!(2k+12)(2k+14)(2k+15)!!}{(k+10)!(k+5)!}&\left(\because \Gamma\left(\frac12\right)=\sqrt{\pi}\right)\\
&=\frac{(2k+10)!!(2k+12)(2k+14)}{(k+10)!(k+5)!}\frac{(2k+15)!}{(2k+14)!!}\\
&=\frac{(2k+15)!}{(k+10)!(k+5)!}\\
\end{align*}
PS: I think that Wolfram Alpha's algorithm doesn't apply the common logic of choosing $k+5$ people from $2k+15$.
