Evaluate this sum: $ \sum_{q=0}^{2n} \binom{p+l-q}{p} \binom{2n}{q}$ In the midst of a calculation, I ran into the following sum.  I'd like to find a form for it which is more explicit, although I haven't figured anything out yet.  Here it is:
Let $p,l$ and $n$ be positive integers with $p+l\geq 2n$.  Then I would like to evaluate:
$$
\sum\limits_{q=0}^{2n}\begin{pmatrix}p+l-q\\p\end{pmatrix}\begin{pmatrix}2n\\q\end{pmatrix}
$$
Any ideas/hints/thoughts?  Thanks!
 A: At least for some special values we can obtain a closed expression. In the following we use the coefficient of operator $[z^p]$ to denote the coefficient of $z^p$ in a series. This way we can write e.g.
\begin{align*}
[z^p](1+z)^q=\binom{q}{p}
\end{align*}

Special case: $p+l=2n$
Since $p+l\geq 2n$ we look at first at $p+l=2n$ and show
  \begin{align*}
\sum_{q=0}^{2n}\binom{2n-q}{p}\binom{2n}{q}=\binom{2n}{p}2^{2n-p}\qquad\qquad\qquad n,p\geq 0
\end{align*}
We obtain
  \begin{align*}
\sum_{q=0}^{2n}\binom{2n-q}{p}\binom{2n}{q}
&=\sum_{q=0}^\infty[u^p](1+u)^{2n-q}[z^q](1+z)^{2n}\tag{1}\\
&=[u^p](1+u)^{2n}\sum_{q=0}^\infty\left(\frac{1}{1+u}\right)^{-q}[z^q](1+z)^{2n}\tag{2}\\
&=[u^p](1+u)^{2n}\left(1+\frac{1}{1+u}\right)^{2n}\tag{3}\\
&=[u^p](2+u)^{2n}\tag{4}\\
&=\binom{2n}{p}2^{2n-p}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we apply the coefficient of operator and extend the upper limit to $\infty$ without changing anything since we are adding zeros only.

*In (2) we use the linearity of the coefficient of operator.

*In (3) we apply the substitution rule of the coefficient of operator with $z=\frac{1}{1+u}$.
\begin{align*}
A(u)=\sum_{q=0}^\infty a_q u^q=\sum_{q=0}^\infty u^q[z^q]A(z)
\end{align*}

*In (4) we do some simplifications and select the coefficient of $u^p$.

Special case: $p+l=2n+1$
We can do the calculation similarly as above and obtain
  \begin{align*}
\sum_{q=0}^{2n}\binom{2n+1-q}{p}\binom{2n}{q}
&=[u^p](1+u)(2+u)^{2n}\tag{5}\\
&=\left([u^p]+[u^{p-1}]\right)(2+u)^{2n}\tag{6}\\
&=\binom{2n}{p}2^{2n-p}+\binom{2n}{p-1}2^{2n-p+1}\tag{7}
\end{align*}

Comment:


*

*In (5) we see the step corresponding to the calculation as we did in (4) with an additional factor $\color{blue}{(1+u)^1}$ corresponding to $p+l=2n\color{blue}{+1}$.

*In (6) we use again the linearity of the coefficient of operator and apply the rule $[u^{p}]u^qA(u)=[u^{p-q}]A(u)$.

*In (7) we select the coefficient of $u^p$ and $u^{p-1}$.
We consider now the general case $p+l\geq 2n$.

General case: $p+l=2n+m\geq 2n\qquad \qquad(m\geq 0)$
We can use the same approach as above and obtain
  \begin{align*}
\sum_{q=0}^{2n}\binom{2n+m-q}{p}\binom{2n}{q}
&=\sum_{q=0}^\infty[u^p](1+u)^{2n+m-q}[z^q](1+z)^{2n}\\
&=[u^p](1+u)^m(2+u)^{2n}\\
&=[u^p]\sum_{j=0}^{m}\binom{m}{j}u^j(2+u)^{2n}\\
&=\sum_{j=0}^{\min\{m,p\}}\binom{m}{j}[u^{p-j}](2+u)^{2n}\\
&=\sum_{j=0}^{\min\{m,p\}}\binom{m}{j}\binom{2n}{m-j}2^{2n-m+j}
\end{align*}

We note the general case does regrettably not provide a simplification.
A: The solution to this involves a hypergeometric function (specifically the Gaussian one):
$$S=\binom{l+p}{p}{}_2F_1\left(-l,-2n;-l-p;-1 \right).$$
There are lots of transformation formulas (for instance here and here), but I doubt you will be able to reduce it to elementary function. 
By Eq. (15.2.4) in the second link, we can write
$$S=\binom{l+p}{p}\sum_{k=0}^l \binom{l}{k}\frac{(-2n)_k}{(-l-p)_k},$$
where Pochhammer symbols have been used (note the remark after the formula, however).
