<s>[INST] prove $\sum_{k = 0}^{n} \binom{n}{k} \binom{m-n}{n-k} = \binom{m}{n}$ prove $\sum_{k = 0}^{n} \binom{n}{k} \binom{m-n}{n-k} = \binom{m}{n}$
Attempt:I was thinking of trying to prove this through induction, but I am having trouble with a base case:
base case:  let $n = 2$: $$LHS = \binom{2}{0} \binom{m-2}{2} + \binom{2}{1} \binom{m-2}{1} + \binom{2}{2} \binom{m-2}{0} \\ = \frac{(m-2)[(m-3) + 2!2!(m-3)!] + 2!}{2!} \ (after\ simplification)$$
$$RHS = \frac{m!}{2!(m-2)!}$$
But I am stuck as what to try next to at least equate these two expressions.
Note: I took a look at Vandermonde's identity on wikipedia and the ensuing proof, but the proof still leaves out how to make the transition between the two initial expressions
 [/INST] Use Pascal’s identity a few times:
$$\begin{align*}
\binom20\binom{m-2}2+\binom21\binom{m-2}1+\binom22\binom{m-2}0&=\binom{m-2}2+2\binom{m-2}1+\binom{m-2}0\\
&=\left(\binom{m-2}2+\binom{m-2}1\right)+\\
&\qquad\qquad\left(\binom{m-2}1+\binom{m-2}0\right)\\
&=\binom{m-1}2+\binom{m-1}1\\
&=\binom{m}2\;.
\end{align*}$$
But why start at $n=2$? Starting at $n=0$ makes the base case very simple.
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