Finding the summation of a product of the particular binomial coefficients: $\sum_{j=0}^{k} \binom{n-j}{p} \binom{m+j}{q}$ How can I simplify the following expression?

$$\sum_{j=0}^{k} \binom{n-j}{p} \binom{m+j}{q}$$

where $n,m,p,q,k$ are positive constants 
such that $n-k \ge p$ and $m \ge q$.
 A: If $n-k=p$ and $m=q$, then we can use the identity
$$
\begin{align}
(1-x)^{-p-1}
&=\sum_{i=0}^\infty(-1)^i\binom{-p-1}{i}x^i\\
&=\sum_{i=0}^\infty\binom{p+i}{i}x^i\\
&=\sum_{i=0}^\infty\binom{p+i}{p}x^i\tag{1}
\end{align}
$$
to get
$$
\begin{align}
(1-x)^{-p-1}(1-x)^{-q-1}
&=\sum_{i=0}^\infty\binom{p+i}{p}x^i\;\;\sum_{j=0}^\infty\binom{q+j}{q}x^j\\
&=\sum_{j=0}^\infty\sum_{i=0}^\infty\binom{p+i}{p}x^i\binom{q+j}{q}x^j\\
&=\sum_{j=0}^\infty\sum_{n=j+p}^\infty\binom{n-j}{p}x^{n-j-p}\binom{q+j}{q}x^j\\
&=\sum_{n=p}^\infty\sum_{j=0}^{n-p}\binom{n-j}{p}\binom{q+j}{q}x^{n-p}\tag{2}
\end{align}
$$
Of course, from $(1)$ we get
$$
(1-x)^{-p-q-2}=\sum_{k=0}^\infty\binom{p+q+1+k}{p+q+1}x^k\tag{3}
$$
Equating the terms with identical powers of $x$ and remembering that $n-k=p$ and $m=q$, we get
$$
\sum_{j=0}^{n-p}\binom{n-j}{p}\binom{m+j}{q}=\binom{n+m+1}{p+q+1}\tag{4}
$$
A: Another hypergeometric expression from Maple:
$$ {n\choose p}{m\choose q}{\mbox{$_3$F$_2$}(1,m+1,-n+p;\,-n,m-q+1;\,1)}-
{n-1-k\choose p}{m+1+k\choose q}
{\mbox{$_3$F$_2$}(1,m+2+k,-n+1+k+p;\,-n+k+1,m+k+2-q;\,1)}
$$
