Sum of binomial coefficients involving $n,p,q,r$ 
Sum of binomial product $\displaystyle \sum^{n}_{r=0}\binom{p}{r}\binom{q}{r}\binom{n+r}{p+q}$

Simplifying $\displaystyle \frac{p!}{r!\cdot (p-r)!} \cdot \frac{q!}{r!\cdot (q-r)!}\cdot \frac{(n+r)!}{(p+q)! \cdot (n+r-p-q)!}$.
Could some help me with this, thanks
 A: It is  convenient  to use the coefficient of operator $[t^r]$ to denote the coefficient of $t^r$  in a series.  This way we can write e.g.
\begin{align*}
\binom{p}{r}=[t^r](1+t)^p
\end{align*}

The following is valid
  \begin{align*}
\sum_{r=0}^n\binom{p}{r}\binom{q}{r}\binom{n+r}{p+q}=\binom{n}{p}\binom{n}{q}
\end{align*}
We obtain
  \begin{align*}
\sum_{r=0}^n&\binom{p}{r}\binom{q}{r}\binom{n+r}{p+q}\\
&=\sum_{r=0}^n\binom{p}{r}\binom{q}{q-r}\binom{n+r}{p+q}\\
&=\sum_{r=0}^\infty[t^r](1+t)^p[v^{q-r}](1+v)^q[w^{p+q}](1+w)^{n+r}\tag{1}\\
&=[v^q](1+v)^q[w^{p+q}](1+w)^n\sum_{r=0}^\infty(v(1+w))^r[t^r](1+t)^p\tag{2}\\
&=[v^q](1+v)^q[w^{p+q}](1+w)^n(1+v(1+w))^p\tag{3}\\
&=[v^q](1+v)^{p+q}[w^{p+q}](1+w)^n\left(1+\frac{vw}{1+v}\right)^p\tag{4}\\
&=[v^q](1+v)^{p+q}[w^{p+q}]\sum_{k=0}^{p+q}\binom{n}{k}w^k\sum_{j=0}^p\binom{p}{j}\left(\frac{v}{1+v}\right)^jw^j\tag{5}\\
&=[v^q](1+v)^{p+q}\sum_{k=0}^{p+q}\binom{n}{k}[w^{p+q-k}]\sum_{j=0}^p\binom{p}{j}\left(\frac{v}{1+v}\right)^jw^j\\
&=[v^q](1+v)^{p+q}\sum_{k=0}^{p+q}\binom{n}{k}\binom{p}{p+q-k}\frac{v^{p+q-k}}{(1+v)^{p+q-k}}\tag{6}\\
&=\sum_{k=p}^{p+q}\binom{n}{k}\binom{p}{k-q}[v^{k-p}](1+v)^k\tag{7}\\
&=\sum_{k=p}^{p+q}\binom{n}{k}\binom{p}{k-q}\binom{k}{p}\\
&=\binom{n}{p}\sum_{k=p}^{p+q}\binom{p}{k-q}\binom{n-p}{k-p}\tag{8}\\
&=\binom{n}{p}\sum_{k=0}^{q}\binom{p}{k+p-q}\binom{n-p}{k}\tag{9}\\
&=\binom{n}{p}\sum_{k=0}^{q}\binom{p}{q-k}\binom{n-p}{k}\tag{10}\\
&=\binom{n}{p}\binom{n}{q}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we apply the coefficient of operator for each binomial coefficient.

*In (2) we use the linearity of the coefficient of operator and apply the rule $[t^{p}]t^qA(t)=[t^{p-q}]A(t)$.

*In (3) we apply the substitution rule of the coefficient of operator with $t:=v(1+w)$
\begin{align*}
A(z)=\sum_{r=0}^\infty a_rz^r=\sum_{r=0}^\infty z^r[t^r]A(t)
\end{align*}

*In (4) we factor out $(1+v)^p$.

*In (5) we apply the binomial summation formula twice. Since we want to select the coefficient of $w^{p+q}$ we can restrict the upper limit of the left sum with $k=p+q$.

*In (6) we select the coefficient of $w^{p+q-k}$.

*In (7) we do some simplifications and can restrict the lower limit of the sum with $k=p$.

*In (8) we apply the cross product $\binom{n}{k}\binom{k}{j}=\binom{n}{j}\binom{n-j}{k-j}$.

*In (9) we shift the summation index and start from $k=0$.

*In (10) we apply the Chu-Vandermonde identity.
A: This is Bizley’s identity (6.42 in Gould's book "Combinatorial identities"):
$$\sum^{n}_{r=0}\binom{p}{r}\binom{q}{r}\binom{n+r}{p+q}=\binom{n}{p}\binom{n}{q}.$$
A nice combinatorial proof can be found in the following paper by Székely as a special case of a more general identity (see Corollary 3):
Common origin of cubic binomial identities. A generalization of Surányi's proof on Le Jen Shoo's formula
A: The proposed identity can be derived (putting $r=s$) from this more general one
$$
\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( \matrix{
  m - \left( {r - s} \right) \cr 
  k \cr}  \right)\left( \matrix{
  n + \left( {r - s} \right) \cr 
  n - k \cr}  \right)\left( \matrix{
  r + k \cr 
  m + n \cr}  \right)}  = \left( \matrix{
  r \cr 
  m \cr}  \right)\left( \matrix{
  s \cr 
  n \cr}  \right)\quad \quad \left| \matrix{
  \,{\rm 0} \le {\rm integer \, }m,n \hfill \cr 
  \;{\rm real}\;r,s \hfill \cr}  \right.
$$
reported in "Concrete Mathematics" - pag. 171, and which is demonstrated through the following passages:
$$
\begin{gathered}
  \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
  m - \left( {r - s} \right) \\ 
  k \\ 
\end{gathered}  \right)\left( \begin{gathered}
  n + \left( {r - s} \right) \\ 
  n - k \\ 
\end{gathered}  \right)\left( \begin{gathered}
  r + k \\ 
  m + n \\ 
\end{gathered}  \right)}  =  \hfill \\
   = \sum\limits_{\begin{subarray}{l} 
  \left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right) \\ 
  \left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,n} \right) 
\end{subarray}}  {\left( \begin{gathered}
  m - \left( {r - s} \right) \\ 
  k \\ 
\end{gathered}  \right)\left( \begin{gathered}
  n + \left( {r - s} \right) \\ 
  n - k \\ 
\end{gathered}  \right)\left( \begin{gathered}
  r \\ 
  m + n - j \\ 
\end{gathered}  \right)\left( \begin{gathered}
  k \\ 
  j \\ 
\end{gathered}  \right)}  =  \hfill \\
   = \sum\limits_{\begin{subarray}{l} 
  \left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right) \\ 
  \left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,m + n} \right) 
\end{subarray}}  {\left( \begin{gathered}
  m - \left( {r - s} \right) \\ 
  k \\ 
\end{gathered}  \right)\left( \begin{gathered}
  k \\ 
  j \\ 
\end{gathered}  \right)\left( \begin{gathered}
  n + \left( {r - s} \right) \\ 
  n - k \\ 
\end{gathered}  \right)\left( \begin{gathered}
  r \\ 
  m + n - j \\ 
\end{gathered}  \right)}  =  \hfill \\
   = \sum\limits_{\begin{subarray}{l} 
  \left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right) \\ 
  \left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,m + n} \right) 
\end{subarray}}  {\left( \begin{gathered}
  m - \left( {r - s} \right) \\ 
  j \\ 
\end{gathered}  \right)\left( \begin{gathered}
  m - \left( {r - s} \right) - j \\ 
  k - j \\ 
\end{gathered}  \right)\left( \begin{gathered}
  n + \left( {r - s} \right) \\ 
  n - k \\ 
\end{gathered}  \right)\left( \begin{gathered}
  r \\ 
  m + n - j \\ 
\end{gathered}  \right)}  =  \hfill \\
   = \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered}
  m - \left( {r - s} \right) \\ 
  j \\ 
\end{gathered}  \right)\left( \begin{gathered}
  m + n - j \\ 
  n - j \\ 
\end{gathered}  \right)\left( \begin{gathered}
  r \\ 
  m + n - j \\ 
\end{gathered}  \right)}  =  \hfill \\
   = \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered}
  m - \left( {r - s} \right) \\ 
  j \\ 
\end{gathered}  \right)\left( \begin{gathered}
  m + n - j \\ 
  m \\ 
\end{gathered}  \right)\left( \begin{gathered}
  r \\ 
  m + n - j \\ 
\end{gathered}  \right)}  =  \hfill \\
   = \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered}
  m - \left( {r - s} \right) \\ 
  j \\ 
\end{gathered}  \right)\left( \begin{gathered}
  r \\ 
  m \\ 
\end{gathered}  \right)\left( \begin{gathered}
  r - m \\ 
  n - j \\ 
\end{gathered}  \right)}  =  \hfill \\
   = \left( \begin{gathered}
  r \\ 
  m \\ 
\end{gathered}  \right)\left( \begin{gathered}
  s \\ 
  n \\ 
\end{gathered}  \right) \hfill \\ 
\end{gathered}
$$
consisting in:  


*

*Vandermonde de-convolution

*shift of the 4th binomial

*"trinomial revision"

*sum on $k$ by Vandermonde convolution

*symmetry

*"trinomial revision"

*sum on $j$ by Vandermonde convolution

