Cousin of the Vandermonde binomial identity The Vandermonde binomial identity can be expressed as
\begin{align*}
\sum_{i+j=r} \binom{m}{i} \binom{n}{j} = \binom{m+n}{r} && r \leq m +n.
\end{align*}
While working on an algebra problem, I stumbled on a formally similar, but distinct identity:
\begin{align*}
\sum_{i+j=r} \binom{i}{m}\binom{j}{n} = \binom{r+1}{m+n+1}
 && m+n \leq r.
\end{align*}
This isn't hard to prove or anything. The left-hand side enumerates the subsets  $S \subseteq \{1,2,\ldots,r+1\}$ with $|S| = m+n+1$ according to the position of the $(m+1)$st largest element of $S$. But, I found the similarity striking enough to ask the following

Question: Are the parallels between these two formulas just a coincidence? Or, is there something else going on here?

 A: This is not a complete explanation but this observation might be helpful:
Consider the sum
$$
S(s,r)= \sum_{m+n=s}\sum_{i+j=r} \binom{m}{i} \binom{n}{j},
$$
with $r\leq s$.
Using the first formula, it becomes
$$
S(s,r) = \sum_{m+n=s} \binom{s}{r} = (s+1) \binom{s}{r} = \frac{(s+1)!}{r!(s-r)!}.
$$
On the other hand, changing the order of summation and using the second fromula, it becomes
$$
S(s,r) = \sum_{i+j=r} \binom{s+1}{r+1} = (r+1) \binom{s+1}{r+1} = \frac{(s+1)!}{r!(s-r)!}.
$$
This suggests that there might be a nice way of calculating $S(s,r)$ (or perhaps a combinatorial interpretation of $S(s,r)$) that shows an intuitive relationship between the formulas you gave.  

There is indeed a combinatorial interpretation:
Consider a set of $s+1$ people.
$$\{1,2,\ldots,s+1\}$$ 
We want to count the number of ways to select one chairperson and $r$ committee members from this set. 


*

*Directly, there are $(s+1) \cdot \binom{s}{r}$ ways to do this. 

*Indirectly, if $x$ is the chairperson, then there are $m$ people to the left of $x$ and $n$ people to the right of $x$ where $m+n=s$, the remaining people who could be on the committee. We now tally up the possible committees conditional on there being $i$ people selected from the left-hand group and $j=r-i$ selected from the right-hand group. 

A: It can be adjusted to be a convolution by shifting the upper and lower indices and then there must be a formal power series $f(x)$ for which the second identity 

equates the $x^t$ terms in $f(x)^p f(x)^q = f(x)^{p+q}$, 

just as the first identity compares the $x^r$ coefficients of  $(1+x)^m (1+x)^n = (1+x)^{m+n}$. 
Taking $(I,J,R,M,N) = (i+1, j+1, r+2, m+1, n+1) $ the formula becomes 
\begin{align*}
\sum_{I+J=R} \binom{I-1}{M-1}\binom{J-1}{N-1} = \binom{R-1}{M+N-1}
 && M+N \leq R.
\end{align*}
which now has the correct form $\sum_{I+J=R} c(M,I)c(N,J) =c(M+N,R)$ 
for $c(p,q)={{q-1} \choose {p-1}}$.   
We can read off $f(x)$ as $\sum c(1,v)x^v = 0 + x + x^2 + x^3 + \dots = \frac{x}{1-x}$ , with $(t,p,q)=(R,M,N)$ and write everything in the original variables. The identity 

equates the $x^{r+2}$ coefficients of  $\hskip10pt (\frac{x}{1-x})^{m+1}(\frac{x}{1-x})^{n+1} = (\frac{x}{1-x})^{m+n+2}.$

Cancelling the powers of $x$ gives a proof using the binomial theorem with negative exponents, by

equating the $x^{r-m-n}$ coefficients of  $\hskip10pt (\frac{1}{1-x})^{m+1}(\frac{1}{1-x})^{n+1} = (\frac{1}{1-x})^{m+n+2}.$

