**Question**: How to calculate this summation $S=\sum\_{k=0}^m a^k b^{m-k} {n\_1\choose k} {n\_2\choose m-k} $? Where $m<n\_1,m<n\_2$


**Remark1**: When $a=b$, I know the above summation $S=a^m\sum\_{k=0}^m {n\_1\choose k} {n\_2\choose m-k} =a^m {n\_1+n\_2\choose m} $.


**Remark2**: This summation looks somewhat similar to the usual binomial formula $\sum\_{k=0}^m a^k b^{m-k} {m\choose k} =(a+b)^m$. So is there also a similar formula for $S$?



How to calculate this summation $\sum_{k=0}^m a^k b^{m-k} {n_1\choose k} {n_2\choose m-k} $?
In terms of a hypergeometric function you would have
$$S=\sum\_{k=0}^m a^k b^{m-k} {n\_1\choose k} {n\_2\choose m-k}=b^m \binom{n\_2}{m} \, \_2F\_1\left(-m,-n\_1;-m+n\_2+1;a/b\right).$$
I don't see a simpler closed-form expression for arbitrary parameters, but if you fix $n\_1$ this does simplify to a simple rational function of $m,n\_2,a/b$.


