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I have just asked the calculation of the following summation see here $$S(a,b,m,n_1,n_2)=\sum_{k=0}^m a^k b^{m-k} {n_1\choose k} {n_2\choose m-k}, $$ which is motivated by the calculation of the ...

An interesting combinatorial identity is the Vandermonde convolution identity: $$ \sum_k {n\choose k}{m\choose s-k} = {n+m \choose s},$$ which can be proved by considering the coefficients in $(x+1)^{...