Pseudo Vandermonde fomula Given integers $m,n$ such that $1\le m\le n$,  let us consider the set $SI_{m,n}$ of all strictly increasing maps from $\{1,\ldots,m\}$ to $\{1,\ldots,n\}$.
It is well known that $SI_{m,n}=\binom nm$.
Now let us choose $i\in\{1,\ldots,m\}$, $j\in\{1,\ldots,n\}$ and consider the maps $f\in SI_{m,n}$ which satisfy $f(i)=j$.
We see that :


*

*if $j<i$ or $j>n-m+i$, then there are no such maps

*if $i\le j\le n-m+i$, there are $\binom{j-1}{i-1}\binom{n-j}{m-i}$ such maps


Summing up, we get the nice formula :
$$\sum_{j=i}^{n-m+i}\binom{j-1}{i-1}\binom{n-j}{m-i}=\binom nm$$
Putting $p=i-1$, $q=j-1$ and re-indexing, we can rewrite it as follows :

$$\sum_{j=p}^{n-q}\binom{j}{p}\binom{n-j}{q}=\binom{n+1}{p+q+1}$$

Wich looks pretty much like but IS NOT the well-known Vandermonde formula.
Hence my question : Does this formula has a standard name ? To whom is it commonly attributed ?
 A: Recall the Chu-Vandermonde identity 
\begin{align*}
\sum_{j=0}^m\binom{s}{j}\binom{t}{m-j}=\binom{s+t}{m}\qquad\qquad s,t\in\mathbb{C}\tag{1}
\end{align*}
is a generalisation of Vandermonde's identity for arguments not only integers but elements from $\mathbb{C}$.

Here we show OPs identity is a Chu-Vandermonde identity in disguise. We obtain
  \begin{align*}
\sum_{j=p}^{n-q}\binom{j}{p}\binom{n-j}{q}&=\sum_{j=0}^{n-p-q}\binom{j+p}{p}\binom{n-j-p}{q}\tag{2}\\
&=\sum_{j=0}^{n-p-q}\binom{j+p}{j}\binom{n-p-j}{n-p-q-j}\tag{3}\\
&=\sum_{j=0}^{n-p-q}\binom{-p-q}{j}\binom{-q-1}{n-p-q-j}(-1)^{n-p-q}\tag{4}\\
&=(-1)^m\sum_{j=0}^m\binom{-p-1}{j}\binom{-q-1}{m-j}\tag{5}\\
&=(-1)^m\binom{-p-q-2}{m}\tag{6}
\end{align*}

Comment:


*

*In (2) we shift the index $j$ to start with $j=0$.

*In (3) we use the binomial identity $\binom{n}{k}=\binom{n}{n-k}$.

*In (4) we use the binomial identity $\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k$.

*In (5) we use the substitution $m=n-p-q$.

*In (6) we apply the Chu-Vandermonde identity (1).

We conclude from (5) and (6) OP's binomial identity is equivalent with the Chu-Vandermonde identity
\begin{align*}
\color{blue}{\sum_{j=0}^m\binom{-p-1}{j}\binom{-q-1}{m-j}=\binom{-p-q-2}{m}}
\end{align*}

A: Identity 5.26 of Concrete Mathematics (Graham, Knuth, and Patashnik) is $$\sum_{0\le k\le l} \binom{l-k}{m} \binom{q+k}{n} = \binom{l+q+1}{m+n+1}$$ Under the substitutions $(k,l,m,n,q) \to (j,n,q,p,0)$ we get $$\sum_{0\le j\le n} \binom{j}{p} \binom{n-j}{q} = \binom{n+1}{p+q+1}$$ and this is the same as your sum because although the indices of the sum differ, in the range where they differ the summand is zero.
Now for the disappointment: this is Vandermonde in disguise. Exercise 5.14 is

Prove identity (5.25) by negating the upper index in Vandermonde's convolution (5.22). Then show that another negation yields (5.26).

For completeness, their presentation of (5.22) is $$\sum_k \binom{r}{m+k} \binom{s}{n-k} = \binom{r+s}{m+n}$$ and (5.25) is $$\sum_{k \le l} \binom{l-k}{m} \binom{s}{k-n} (-1)^k = (-1)^{l+m} \binom{s-m-1}{l-m-n}$$
GKP give no name to the identity nor attribution for it, and since they're usually good with attribution I assume that none is known.
