There is a nice combinatorial interpretation. The product $\binom{x}{i}\cdot\binom{x}{j}$ gives the number of ways in which one can select $i$ objects among $x$, then select $j$ objects among $x$. Let $I$ be the set of the elements chosen in the first instance, $J$ be the set of the elements chosen in the second one. In how many cases $I$ and $J$ are disjoint? We have to choose $i+j$ elements among $x$, then choose the elements of $I$ among the $i+j$ selected, so $I$ and $J$ are disjoint in: $\binom{x}{i+j}\binom{i+j}{i}$ cases. In how many cases we have $|I\cap J|=k\leq i$? The answer is clearly: $\binom{x}{i+j-k}\binom{i+j-k}{k}\binom{i+j-2k}{i-k},$ since we have to choose the $(i+j-k)$ elements that belong to $I\cup J$, which $k$ of these belong to $I\cap J$, which $i-k$ of the remaining belong to $I$. So we have: $\binom{x}{i}\cdot\binom{x}{j}=\sum_{k=0}^{i}\binom{x}{i+j-k}\binom{i+j-k}{k}\binom{i+j-2k}{i-k}.$