Vandermonde's identity check in probability The book on probability I'm reading states Vandermonde's identity as: $\binom{m+n}{k} = \sum_{i=0}^k \binom{m}{i} \binom{n}{k-i}$, but further in the book I'm seeing it being used as: $\binom{m+n}{k} = \sum_{i=0}^n \binom{m}{k-i} \binom{n}{i}$ and I can't seem to show that these are the same.  Is this identity wrong or am I missing something?

The identity is correct both ways. All that matters is that the index run over all non-zero products of the two binomials. In the first version we know that $\binom{n}{k-i}$ is $0$ if $i<0$ or $i>k$, so having $i$ run from $0$ to $k$, inclusive, ensures that we include all of the non-zero terms. We may also include some $0$ terms, but that does no harm. (This happens if $k>m$.) In the second we know that $\binom{n}i$ is non-zero precisely when $0\le i\le n$, so here again the range of summation ensures that we get all of the non-zero terms. The two versions could simply be written
$$\binom{m+n}k=\sum_i\binom{m}i\binom{n}{k-i}$$
and
$$\binom{m+n}k=\sum_i\binom{m}{k-i}\binom{n}i\;,$$
with $i$ ranging over all integers, since only finitely many terms are non-zero. This is why it’s convenient to adopt the convention that $\binom{n}k=0$ if $k<0$ or $k>n$.
