I need to show that $$\sum_{k}\binom{n+k-1}{n+\alpha-1}\binom{\beta-k-1}{n-k} = \binom{n+\beta-1}{n-\alpha}$$.
Proof: Putting $n-k = m$, L.H.S becomes, \begin{align*} & \sum_{m}\binom{2n-m-1}{n+\alpha-1}\binom{\beta-n+m-1}{m}\\ &= \sum_{m} \binom{2n-m-1}{n-m-\alpha}\binom{\beta-n+m-1}{m} \quad \quad \text{by symmetry}\\ &= \sum_{m} \binom{2n-m-1}{(n-\alpha)-m}\binom{\beta-n+m-1}{m}\\ &= \binom{n+\beta-2}{n-\alpha} \quad \quad \text{by Vandermonde convolution} \end{align*}
Note: 1) Symmetry says: $$\binom{n}{r} = \binom{n}{n-r}$$ 2) Vandermonde convolution says $$\sum_k \binom{r}{k}\binom{s}{n-k}=\binom{r+s}{n}$$
Kindly someone tell, where am I getting wrong. Thanks.
Vandermonde does not involve the summation index in the top of the binomial coefficients. If it did, you could more simply obtain the same false result by first replacing $\binom{n+k-1}{n-\alpha-1}$ with $\binom{n+k-1}{k-\alpha}$.