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}$?

Question

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 $q=1-p$.

Answer 1

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.