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?
 A: 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 $k>n$.
A: The only difference between two two identities is in interchanging $m$ and $n$, the order of the two binomial components being multiplied, and finally a difference in the upper bound of the summation. Applying the first two changes to the second identity, we are now comparing
$$
 \binom{m+n}{k} = \sum_{i=0}^k \binom{m}{i} \binom{n}{k-i}
\quad\hbox{to}\quad
 \binom{n+m}{k} = \sum_{i=0}^m  \binom{m}{i}\binom{n}{k-i}
$$
The first summation stops where the binomial coefficient $\binom n{k-i}$ would become zero because of a negative lower index if one would go on, while the second one stops where the binomial coefficient $\binom mi$ would become zero because of a too large lower index if one would go on. The summand becomes zero beyond the limit in either case, so no terms that are in one summation but not the other make any difference.
One could alternatively limit the summation to $\max(0,k-n)\leq i\leq\min(m,k)$ to consider only nonzero terms, or go to the other extreme of summing over $i\in\Bbb Z$ (with all but finitely many terms zero) to stress the fact that no limits are affecting the summation at all (I prefer the latter appraoch).
