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*}
Best Answer
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 $$ \sum_{i_1+\cdots+i_n=k}\binom{r_1+i_1}{r_1}\cdots\binom{r_n+i_n}{r_n}=[x^k]\frac{1}{(1-x)^{r_1+\cdots+r_n+n}}={r_1+\cdots+r_n+n-1+k \choose k} $$