Evaluate $\sum_{k=0}^{n} {n \choose k}{m \choose k}$ for a given $n$ and $m$.

Use the fact that (which follows from the definition) $$\binom{m}{k}=\binom{m}{m-k}.$$ Once you have this the LHS can be written as $$LHS=\sum_{k=0}^n\binom{n}{k}\binom{m}{k}=\sum_{k=0}^n\binom{n}{k}\binom{m}{m-k}.$$

Now we can do a combinatorial argument to find this sum. Consider a group of $n$ men and $m$ women. We want to make a committee consisting of $m$ people. This can be done in any of the following ways:

1). $0$ men and $m$ women-----this selection can be made in $\binom{n}{0}\binom{m}{m}$ ways.

2). $1$ man and $m-1$ women-----this selection can be made in $\binom{n}{1}\binom{m}{m-1}$ ways.

and so on.....

m+1). $m$ men and $0$ women-----this selection can be made in $\binom{n}{m}\binom{m}{0}$ ways.

The sum total of this gives you the LHS. But this problem can also be solved by considering choosing $m$ people out of a group of $m+n$ people, which can be done in $\binom{n+m}{m}$ ways. Hence the two ways of counting should be equal.

$$\sum_{k=0}^n\binom{n}{k}\binom{m}{m-k}=\binom{n+m}{n}=\binom{n+m}{m}$$

Using the Vandermonde Identity, this is $$ \sum_{k=0}^n\binom{n}{n-k}\binom{m}{k}=\binom{n+m}{n} $$

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Proof of Vandermonde's Identity

$$ \begin{align} \sum_{k=0}^{m+n}\color{#C00000}{\binom{m+n}{k}}x^k &=(1+x)^{m+n}\tag{1}\\ &=(1+x)^m(1+x)^n\tag{2}\\ &=\sum_{j=0}^m\binom{m}{j}x^j\sum_{k=0}^n\binom{n}{k}x^k\tag{3}\\ &=\sum_{j=0}^m\sum_{k=j}^{n+j}\binom{m}{j}\binom{n}{k-j}x^k\tag{4}\\ &=\sum_{k=0}^{m+n}\color{#C00000}{\sum_{j=0}^k\binom{m}{j}\binom{n}{k-j}}x^k\tag{5} \end{align} $$ Explanation:
$(1)$: binomial theorem
$(2)$: property of exponents
$(3)$: binomial theorem
$(4)$: substitute $k\mapsto k-j$
$(5)$: change order of summation

Compare the coefficients of $x^k$.

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${\large\tt\mbox{Hereafter, I'll illustrate a general method:}}$

\begin{align}&\color{#66f}{\large\sum_{k=0}^{n}{n \choose k}{m \choose k}} =\sum_{k=0}^{n}{n \choose k}\oint_{\verts{z}\ =\ 1} {\pars{1 + z}^{m} \over z^{k + 1}}\,{\dd z \over 2\pi\ic} \\[5mm] = &\ \oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{m} \over z}\, \sum_{k=0}^{n}{n \choose k}\pars{1 \over z}^{k}\,{\dd z \over 2\pi\ic} \\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{m} \over z}\, \pars{1 + {1 \over z}}^{n}\,{\dd z \over 2\pi\ic} \\[5mm] = &\ \oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{m + n} \over z^{n + 1}}\, \,{\dd z \over 2\pi\ic} = \color{#66f}{\large{m + n \choose n}} \end{align}