Proving $\sum_i \binom{k}{k-i} \binom{n-k}{i} = \binom{n}{k}$ I am in the middle of a probability question. The question is indeed simple. For the sake of clarity of the notation, I also include the question here, which is from Sheldon M. Ross's Introduction to Probability Models:

A fair coin is independently flipped $n$ times, $k$ times by A and $n-k$ times by B. Show that the probability that A and B flip the same number of heads is equal to the probability that there are a total of $k$ heads.

I have reduced the question to proving $$\sum_i \binom{k}{k-i} \binom{n-k}{i} = \binom{n}{k}$$ and I cannot move on. So far, I have tried to expand $\binom{k}{k-i}$ and $\binom{n-k}{i}$ by the definition of binomial coefficient and then observe for common terms. However, it comes out to be a mess and I am not sure if there is other way to do it.
Thanks in advance.
 A: If in a box there are $n$ marbles, $k$ of them being red and $n-k$ being black, with
$$\sum_{i}\binom{n-k}{i}\binom{k}{k-i}$$
you are counting the ways to select $i$ black marbles and $k-i$ red marbles, for any possible $i$. 
This is just the number of ways to select $k$ marbles out of $n$, i.e. $\binom{n}{k}$.
A: It is simply a matter of considering the coefficient of $x^{k}$ on both sides of
$$
(1 + x)^{k} (1 + x)^{n-k} = (1 + x)^{n}.
$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{}$
\begin{align}
\color{#66f}{\large\sum_{i}{k \choose k - i} {n - k \choose i}}& =
\sum_{i}{n - k \choose i}\bracks{z^{k - i}}\pars{1 + z}^{k} =
\bracks{z^{k}}\pars{1 + z}^{k}\sum_{i}{n - k \choose i}z^{i}
\\[5mm] & =
\bracks{z^{k}}\pars{1 + z}^{k}\,\pars{1 + z}^{n - k} =
\bracks{z^{k}}\pars{1 + z}^{n} =
\bbox[8px,border:1px groove navy]{\ds{n \choose k}}
\end{align}
