Identity for a sum of product of binomial coefficients For some fixed positive integers $r_1,\ldots,r_n$, I would like to find a sum:
$$
\sum_{i_1+\cdots+i_n=k}\binom{r_1+i_1}{r_1}\cdots\binom{r_n+i_n}{r_n}=\sum_{i_1+\cdots+i_n=k}\binom{r_1+i_1}{i_1}\cdots\binom{r_n+i_n}{i_n},
$$
where $k=0,\ldots,r_1+\cdots+r_n$ ($i_j$ ranges from $0$ to $r_j$, for $j=1,\ldots,n$).
If reformulate the problem. Multiply $n$ finite sums:
$$
\sum_{i_1=0}^{r_1}\binom{r_1+i_1}{r_1}\cdots\sum_{i_n=0}^{r_n}\binom{r_n+i_n}{r_n}
$$
collect and sum parts such that $i_1+\cdots+i_n=k$. What is the result of every such sum.
I have found similar question here, but I can not connect it to this problem. Also found a paper which uses probabilistic method to establish several generalisations of Vandermonde identity (which to my dilettante view is somewhat similar to my problem).
Here is a small example just to be clear what I want to achieve. Let $n=3$ and $r_1=1$, $r_2=2$, $r_3=3$. Now take $k=3$, it takes six combinations of $(i_1,i_2,i_3)$: $(1,1,1)$, $(1,2,0)$, $(1,0,2)$, $(0,1,2)$, $(0,2,1)$, $(0,0,3)$ so that $i_1+i_2+i_3=k$ (note that $i_1, i_2, i_3$ can take values at most $1$, $2$ and $3$ respectively). So the sum is:
\begin{align*}
&&{2\choose1}{3\choose2}{4\choose3}+{2\choose1}{4\choose2}{3\choose3}+{2\choose1}{2\choose2}{5\choose3}+\\
&&{1\choose1}{3\choose2}{5\choose3}+{1\choose1}{4\choose2}{4\choose3}+{1\choose1}{2\choose2}{6\choose3}=\\
&&24+12+20+30+24+20=130.
\end{align*}
 A: Here is what can be obtained with generating function technique:
${r+i \choose r}=[x^i]\frac{1}{(1-x)^{r+1}}$, where $[x^i]f(x)$ is the coefficient of $x^i$ in the power series expansion of $f(x)$. Then
$$
\sum_{i_1+\cdots+i_n=k}\binom{r_1+i_1}{r_1}\cdots\binom{r_n+i_n}{r_n}=[x^k]\frac{1}{(1-x)^{r_1+\cdots+r_n+n}}={r_1+\cdots+r_n+n-1+k \choose k}
$$
A: So if I properly understood your question, you are looking for
$$
S({\bf r}_{\,n} ,m)
 = \sum\limits_{\left\{ {\matrix{   {0\, \le \,k_{\,j} \, \le \,r_{\,j} }  \cr 
   {\,k_{\,1}  + k_{\,2}  +  \cdots  + k_{\,n} \, = \,m}  \cr  } } \right.}
 {\left( \matrix{  r_{\,1}  + k_{\,1}  \cr   k_{\,1}  \cr}  \right)
 \left( \matrix{  r_{\,2}  + k_{\,2}  \cr   k_{\,2}  \cr}  \right) \cdots
 \left( \matrix{  r_{\,n}  + k_{\,n}  \cr   k_{\,n}  \cr}  \right)} 
$$
where we can consider ${\bf r}_{\,n}$ as a vector in $n$ dimensions.
If it was not for the limitation $0\, \le \,k_{\,j} \, \le \,r_{\,j} $ the above would be a convolution of the $n$ binomials, giving
an ogf  which is the product of $1/(1-x)^{r_{\,j}+1 }$ as per Renè answer.
With the excursion of $k_{\,j} $ limited to $[0, \,r_{\,j}]$ we have instead a truncated version of the above which we can express as follows.
$$
\eqalign{
  & F(x,r_{\,j} ) = \sum\limits_{0\, \le \,k\, \le \,r_{\,j} }
 {\left( \matrix{  r_{\,j}  + k \cr   k \cr}  \right)x^{\,k} }
  = \sum\limits_{0\, \le \,k\, \le \,r_{\,j} }
 {\left( \matrix{  r_{\,j}  + k \cr   r_{\,j}  \cr}  \right)x^{\,k} }  =   \cr 
  &  = \sum\limits_{0\, \le \,k\,}
 {\left( \matrix{  r_{\,j}  + k \cr   r_{\,j}  \cr}  \right)x^{\,k} }  - x^{\,r_{\,j}  + 1} \sum\limits_{\,0\, \le \,k\,}
 {\left( \matrix{  2r_{\,j}  + 1 + k \cr   r_{\,j}  \cr}  \right)x^{\,k} }  \cr} 
$$
Indicating with $t_{\,k}$ the coefficient in the second sum
$$
t_{\,k}  = \left( \matrix{  2r_{\,j}  + 1 + k \cr   r_{\,j}  \cr}  \right)
 = {{\left( {2r_{\,j}  + 1 + k} \right)^{\,\underline {\,r_{\,j} \;} } } \over {r_{\,j} !}}
 = {{\left( {r_{\,j}  + 2 + k} \right)^{\,\overline {\,r_{\,j} \,} } } \over {1^{\,\overline {\,r_{\,j} \,} } }}
$$
we have
$$
\eqalign{
  & t_{\,0}  = \left( \matrix{  2r_{\,j}  + 1 \cr   r_{\,j}  \cr}  \right)  \cr 
  & {{t_{\,k + 1} } \over {t_{\,k} }}
 = {{\left( {r_{\,j}  + 3 + k} \right)^{\,\overline {\,r_{\,j} \,} } }
 \over {\left( {r_{\,j}  + 2 + k} \right)^{\,\overline {\,r_{\,j} \,} } }}
 = {{\left( {2r_{\,j}  + 2 + k} \right)} \over {\left( {r_{\,j}  + 2 + k} \right)}} \cr} 
$$
so one way to express a single term would be through the hypergeometric function
$$
\eqalign{
  & F(x,r_{\,j} ) = \sum\limits_{0\, \le \,k\, \le \,r_{\,j} }
 {\left( \matrix{  r_{\,j}  + k \cr   k \cr}  \right)x^{\,k} }
  = \sum\limits_{0\, \le \,k\, \le \,r_{\,j} }
 {\left( \matrix{  r_{\,j}  + k \cr  r_{\,j}  \cr}  \right)x^{\,k} }  =   \cr 
  &  = \sum\limits_{0\, \le \,k\,}
 {\left( \matrix{  r_{\,j}  + k \cr   k \cr}  \right)x^{\,k} }  - x^{\,r_{\,j}  + 1} \sum\limits_{\,0\, \le \,k\,}
 {\left( \matrix{  2r_{\,j}  + 1 + k \cr  r_{\,j}  \cr}  \right)x^{\,k} }  =   \cr 
  &  = {1 \over {\left( {1 - x} \right)^{\,r_{\,j}  + 1} }} - x^{\,r_{\,j}  + 1}
 \left( \matrix{  2r_{\,j}  + 1 \cr   r_{\,j}  \cr}  \right)
{}_2F_{\,1} \left( {\left. {\matrix{   {2r_{\,j}  + 2,\;1}  \cr   {r_{\,j}  + 2}  \cr } \;} \right|\;x} \right) \cr} 
$$
But now, multiplying the terms
$$
G(x,{\bf r}_{\,n} ) = \sum\limits_{0\, \le \,m} {S({\bf r}_{\,n} ,m)x^{\,m} }  = \prod\limits_{j = 1}^n {F(x,r_{\,j} )} 
$$
leads to a complicated expression.
Conclusion: lacking a "compact" form to express the truncated binomial, tere is not much to do to
render similarly "compact" your sum.
