TITLE: Is this combinatorial identity a special case of Saalschutz's theroem?
QUESTION [8 upvotes]: When I solved a question, the following combinatorial identity was used
$$
\sum_{k=0}^{n}(-1)^k{n\choose k}{n+k\choose k}{k\choose j}=(-1)^n{n\choose j}{n+j\choose j}.
$$
But to prove this identity is difficulty for me. I think it may be a special case of the Saalschutz's theroem, wchich is stated as follows
$$
\sum_{k=0}^{n}\frac{(-n)_k(a)_k(b)_k}{(1)_k(c)_k(1+a+b-c-n)_k}=\frac{(c-a)_n(c-b)_n}{(c)_n(c-a-b)_n},
$$
where $(a)_k=a\cdot (a+1)\cdots (a+k-1)$. If it isn't the special case of Saalschutz's theroem, can you present a proof of this identity? All hints, proofs and references are welcome.

REPLY [5 votes]: The result is more easily proved without the use of Saalschutz' theorem.
A good starting point for evaluating such sums may be to write them out in terms of factorials:
$$
(-1)^k \binom{n}{k}\binom{n+k}{k}\binom{k}{j}
=\frac{(-1)^k\,n!\,(n+k)!\,k!}{k!\,(n-k)!\,k!\,n!\,j!\,(k-j)!}
=\frac{(-1)^k\,(n+k)!}{k!\,(n-k)!\,j!\,(k-j)!}
$$
for $j\le k\le n$ (otherwise the LHS is zero). I often think of the sum as over all $k\in\mathbb{Z}$, but with the convention that $1/r!=0$ when $r$ is a negative integer.
You may note that $k$ appears in four of the factorials. This means we can try to rewrite the expression so that we may use the rule
$$
\sum_k\binom{a}{p+k}\binom{b}{q-k}=\binom{a+b}{p+q}
$$
where $a$ or $b$ may be negative using the conversion
$$
\binom{-(m+1)}{k}
=(-1)^k\binom{m+k}{k}
=(-1)^k\binom{m+k}{m}.
$$
We may do so as
$$
\frac{(-1)^k\,(n+k)!}{k!\,(n-k)!\,j!\,(k-j)!}
=(-1)^k\binom{n+k}{k}\binom{n-j}{n-k}\binom{n}{j}
=\binom{-(n+1)}{k}\binom{n-j}{n-k}\binom{n}{j}
$$
which gives us the sum
$$
\sum_k\binom{-(n+1)}{k}\binom{n-j}{n-k}\binom{n}{j}
=\binom{-(j+1)}{n}\binom{n}{j}
=(-1)^n\binom{n+j}{j}\binom{n}{j}.
$$

As Darij pointed out, the result could be obtain a little more directly (without writing out all the factorials) by using
$$
\binom{n}{k}\binom{k}{j}=\binom{n}{j}\binom{n-j}{n-k}.
$$
Of course, since I never remember these kinds of rules, I personally have to write this out as factorials anyway, but doing so only for these two binomials is a bit simpler than doing it for all three (and more so in more complicated cases).