Simplification of Combinatorial Expression I have been solving a problem and have been stuck on it for hours now.
I can't recall any combinatorial approach/formula for the below expression :
$$\sum_{i=a+1}^n {i-1 \choose a}{n-i \choose k-a}$$
Any sort of help would be appreciated. I know it is a norm to share my approach as well to show that I have put an effort but honestly I don't have any clue on this particular question.
Thanks!
 A: What follows is admittedly not the simplest proof. We post here by way
of enrichment and to showcase four techniques, coefficient extractors,
the Iverson bracket, residues and the Leibniz rule. 
We seek to show that
$$\sum_{q=a+1}^n {q-1\choose a} {n-q\choose k-a}
= {n\choose k+1}$$
where $k\ge a$ for the binomial coefficient to be defined,
and $n\ge a+1$ or alternatively
$$\sum_{q=0}^{n-a-1} {q+a\choose a} {n-a-1-q\choose k-a}
= {n\choose k+1}.$$
The LHS is
$$[z^{k-a}] (1+z)^{n-a-1}
\sum_{q\ge 0} {q+a\choose a} (1+z)^{-q} [[q\le n-a-1]]
\\ = [z^{k-a}] (1+z)^{n-a-1}
\sum_{q\ge 0} {q+a\choose a} (1+z)^{-q}
[w^{n-a-1}] \frac{w^q}{1-w}
\\ = [z^{k-a}] (1+z)^{n-a-1} [w^{n-a-1}] \frac{1}{1-w}
\sum_{q\ge 0} {q+a\choose a} (1+z)^{-q} w^q
\\ = [z^{k-a}] (1+z)^{n-a-1} [w^{n-a-1}] \frac{1}{1-w}
\frac{1}{(1-w/(1+z))^{a+1}}
\\ = [z^{k-a}] (1+z)^{n} [w^{n-a-1}] \frac{1}{1-w}
\frac{1}{(1+z-w)^{a+1}}.$$
This is
$$[z^{k-a}] (1+z)^n (-1)^a
\mathrm{Res}_{w=0} \frac{1}{w^{n-a}}
\frac{1}{w-1} \frac{1}{(w-(1+z))^{a+1}}.$$
Now the residue at infinity for $w$ is zero by inspection, residues
sum to zero and the residue at $w=1$ yields
$$[z^{k-a}] (1+z)^n (-1)^a \frac{1}{(-1)^{a+1} z^{a+1}}
= - {n\choose k+1}.$$
This is the claim if we can show that the contribution from the pole at
$w=1+z$ is zero. We get (Leibniz rule)
$$\frac{1}{a!} \left(\frac{1}{w^{n-a}} \frac{1}{w-1}\right)^{(a)}
= \frac{1}{a!}
\sum_{q=0}^a {a\choose q}
\frac{(-1)^q (n-1-a+q)!}{(n-1-a)! \times w^{n-a+q}}
\frac{(-1)^{a-q} (a-q)!}{(w-1)^{a+1-q}}
\\ = (-1)^a \sum_{q=0}^a {n-1-a+q\choose q}
\frac{1}{w^{n-a+q}} \frac{1}{(w-1)^{a+1-q}}.$$
We thus obtain for the contribution
$$[z^{k-a}] (1+z)^n  \sum_{q=0}^a {n-1-a+q\choose q}
\frac{1}{(1+z)^{n-a+q}} \frac{1}{z^{a+1-q}}
\\ =   \sum_{q=0}^a {n-1-a+q\choose q}
[z^{k+1-q}] (1+z)^{a-q} = 0$$
because $a\ge q$ and $k+1\gt a.$ This concludes the argument.
A: 
We obtain
\begin{align*}
\color{blue}{\sum_{i=a+1}^n}&\color{blue}{\binom{i-1}{a}\binom{n-i}{k-a}}\\
&=\sum_{i=a+1}^n\binom{i-1}{i-a-1}\binom{n-i}{n-i-k+a}\tag{1}\\
&=\sum_{i=a+1}^n\binom{-a-1}{i-a-1}(-1)^{i-a-1}\binom{-k+a-1}{n-i-k+a}(-1)^{n-i-k+a}\tag{2}\\
&=(-1)^{n-k-1}\sum_{i=0}^{n-a-1}\binom{-a-1}{i}\binom{-k+a-1}{n-k-1-i}\tag{3}\\
&=(-1)^{n-k-1}\binom{-k-2}{n-k-1}\tag{4}\\
&=\binom{n}{n-k-1}\tag{5}\\
&\,\,\color{blue}{=\binom{n}{k+1}}
\end{align*}
and the claim follows.

Comment:

*

*In (1) we use $\binom{p}{q}=\binom{p}{p-q}$ twice.


*In (2) we apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.


*In (3) we shift the index to start with $i=0$.


*In (4) we apply the Chu-Vandermonde identity.
Here we use that the upper index is in fact $n-k-1$, since $k\geq a$.


*In (5) we apply the identity as we did in (2) again.
A: Consider binary words of length $n$ with $k+1$ ones ... $ \binom{n}{k+1}$.
Let the $(a+1)^{th}$ one occur at the $i^{th}$ position. There are $a$ ones in the $i-1$ positions before this entry ... $\binom{i-1}{a}$. And there are $k-a$ ones in the $n-i$ positions after ... $\binom{n-i}{k-a}$. Now $i$ can vary over $a+1$ and $n$, thus we have
\begin{eqnarray*}
\sum_{i=a+1}^n \binom{i-1}{a} \binom{n-i}{k-a} = \binom{n}{k+1}.
\end{eqnarray*}
I would be interested to see an algebriac proof of this.
Edit (In light of MS's answer below): Use the negative binomial trick twice
\begin{eqnarray*}
\binom{-p}{q}=(-1)^q\binom{p+q-1}{q}.
\end{eqnarray*}
The Vandermonde identity can now be applied, negative binomial again and the result follows.
A: Hint: Try $\binom{n}{k+1}$ this is like Hockey Stick and Vandermonde together. Try to combine each of their combinatorial descriptions. Remember that the Hockey Stick identity is like $\sum _{k=0}^n\binom{k}{\ell}=\binom{n+1}{\ell +1}$ and what it does is Fix the biggest element and pick the remaining $k.$ Here you are picking a middle element! (Call it $i$ and picking to the left and to the right(like on Vandermonde)).
Not serious hint: I have called this identity: "Vandermonde playing Hockey" somewhere before.
