How can I "calculate" $ \sum_{x=0}^{z} {n_1 \choose z-x} {n_2 \choose x} p^z\cdot q^{(n_1+n_2)-z}$? How can I prove that:
$$ \sum_{x=0}^{z} {n_1 \choose z-x}  {n_2 \choose x} p^z\cdot q^{(n_1+n_2)-z} = {n_1+n_2 \choose z}\cdot p^z\cdot q^{(n_1+n_2)-z} $$
Note: $0 \lt q,p \lt 1$,  and $q=1-p$.
 A: At first we note that the sum does not depend on $p$ or $q$. So, we only need to show the Chu-Vandermonde Identity for non-negative integer $z$ and $n_1,n_2\in\mathbb{C}$:
\begin{align*}
  \sum_{x=0}^z\binom{n_1}{z-x}\binom{n_2}{x}=\binom{n_1+n_2}{z}
  \end{align*}
It's convenient to use the  coefficient of operator $[s^z]$ to denote the coefficient of $s^z$ of a series. This way we can write for instance
$$\binom{n}{z}=[s^z](1+s)^n$$

We obtain
  \begin{align*}
\color{blue}{\sum_{x=0}^z}\color{blue}{\binom{n_1}{z-x}\binom{n_2}{x}}
&=\sum_{x\geq 0}[s^{z-x}](1+s)^{n_1}[t^x](1+t)^{n_2}\tag{1}\\
&=[s^z](1+s)^{n_1}\sum_{x\geq 0} s^x[t^x](1+t)^{n_2}\tag{2}\\
&=[s^z](1+s)^{n_1}(1+s)^{n_2}\tag{3}\\
&=[s^z](1+s)^{n_1+n_2}\\
&\,\,\color{blue}{=\binom{n_1+n_2}{z}}\tag{4}
\end{align*}
and the claim follows.

Comment:


*

*In (1) we use the coefficient of operator twice and extend the index range to $\infty$ without changing anything since we are adding zeros only.

*In (2) we use the linearity of the coefficient of operator and apply the rule $[s^{a-b}]A(s)=[s^a]s^bA(s)$.

*In (3) we apply the substitution rule of the coefficient of operator with $s=t$
\begin{align*}
A(s)=\sum_{j\geq 0} a_j s^j=\sum_{j\geq 0} s^j [t^j]A(t)
\end{align*}

*In (4) we select the coefficient of $s^z$ accordingly.
