Can $\sum_{i=a}^{b}p^{i}\binom{i}{a}\binom{b}{i}$ be simplified? I'm trying to compute a complicated thing, and I end up with terms like:
$$\sum_{i=a}^{b}p^{i}\binom{i}{a}\binom{b}{i}$$
$a$ and $b$ are nonnegative integers, $0<p<1$.
I don't see a way to make this simpler. Am I missing something or is this the simplest thing that I'm going to get?
 A: Yes, it can be simplified. Note that for $0\leq a\leq b$,
$$\binom{i}{a}\binom{b}{i}=\binom{b}{a}\binom{b-a}{i-a}.$$
Hence
$$\sum_{i=a}^{b}p^{i}\binom{i}{a}\binom{b}{i}=p^a\binom{b}{a}\sum_{i=a}^{b}\binom{b-a}{i-a}p^{i-a}.$$
Can you take it from here?
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\sum_{i = a}^{b}p^{i}{i \choose a}{b \choose i} & =
\sum_{i = 0}^{b}{b \choose i}p^{i}\bracks{z^{a}}\pars{1 + z}^{i} =
\bracks{z^{a}}\sum_{i = 0}^{b}{b \choose i}\pars{p + pz}^{i} \\[5mm] & =
\bracks{z^{a}}\bracks{1 + \pars{p + pz}}^{\, b}
\\[5mm] & =
\pars{1 + p}^{b}\bracks{z^{a}}\bracks{1 + {p \over 1 + p}\,z}^{\, b} \\[5mm] &
\pars{1 + p}^{b}{b \choose a}\pars{p \over 1 + p}^{a}
\\[5mm] & =
\bbx{{p^{a} \over \pars{1 + p}^{a - b}}\,{b \choose a}} \\ &
\end{align}
