\begin{align*} \sum_{x=0}^n \binom{n}{x} y^x & = (1+y)^n, \text{ by the Binomial Theorem}\\ & = (1+y)^{n_1} \cdot (1+y)^{n_2} \\ & = \sum_{x=0}^{n_1} \binom{n_1}{x} y^x \cdot \sum_{x=0}^{n_2} \binom{n_2}{x} y^x \\ & = \sum_{x=0}^n \sum_{t=0}^r \binom{n_1}{r} \binom{n_2}{r-t} y^x \end{align*}