TITLE: Help with binomial identity
QUESTION [11 upvotes]: In my work, I was led to the following identity.  I would be grateful if someone could provide an easy proof.
Suppose $n, d, k \in \mathbb{Z}$, and $d \geq 0$.
$$
\sum_{j = 0}^d (-1)^{d-j} \cdot \binom{d}{k} \cdot \binom{n-j-1}{n-d-1} \cdot \binom{n-d}{j-k} = \left\{
     \begin{array}{ll}
       0 & : d \neq k\\
       0 & : n < 0\\
       1 & : 0 \leq n \leq d\\
       2 & : d < n
     \end{array}
   \right.
$$
I've been to the wikipedia page on binomial coefficients, looking for useful identities to apply.  https://en.wikipedia.org/wiki/Binomial_coefficient
Many well-known identities seem applicable, but I don't know how to handle the way $j$ appears once in the top of one of the binomial coefficients, and once in the bottom.  The identity 
$$
\sum_{k=q}^n \binom{n}{k} \binom{k}{q} = 2^{n-q}\binom{n}{q}
$$
has the moving variable in both places, but I do not see how to use this.
What I'm trying to avoid is proof by induction.  The context is, I need this identity for an example in a paper.  I don't want to get hung up on the example, but I'd like to include a full proof.

REPLY [5 votes]: Substitute $r=d-j$ and use upper negation followed by Vandermonde's Identity:
$$\require{cancel}\begin{align}\sum_{j=0}^d(-1)^{d-j}\binom{d}{k} \binom{n-j-1}{n-d-1} \binom{n-d}{j-k}
&=\binom dk \sum_{r=0}^d(-1)^r\color{blue}{\binom{n-d-1+r}{n-d-1}}\binom{n-d}{d-k-r}\\
&=\binom dk \sum_{r=0}^d(-1)^r\color{blue}{\binom{n-d-1+r}r}\binom{n-d}{d-k-r}\\
&=\binom dk \sum_{r=0}^d(-1)^r\color{blue}{\binom{d-n}r(-1)^r}\binom{n-d}{d-k-r}\\
&=\binom dk \sum_{r=0}^d\binom{d-n}r\binom{n-d}{d-k-r}\\
&=\binom dk \binom 0{d-k}\\
&=\begin{cases}1\quad \text{if}\quad d=k \\0\quad \text{otherwise}\end{cases}
\end{align}\qquad$$
as $$\binom 0{d-k}=\begin{cases}1\quad \text{for}\quad d-k=0\\ 0\quad\text{otherwise}\end{cases}$$