A combinatorial identity: $\sum_{k=0}^j (-1)^k \frac{n-2k}{n-j-k} \binom{n}{k} \binom{n-j-k}{j-k} = 0$ I would like to prove the following identity: For
$0< j\leq \left\lfloor{n \over 2}\right\rfloor$, $$\sum_{k=0}^j (-1)^k \frac{n-2k}{n-j-k} \binom{n}{k} \binom{n-j-k}{j-k} = 0.$$ I have checked for several small $j$, but no clue on how to prove it for all $0<j\leq \lfloor \frac{n}{2} \rfloor$. I have thought about whether one can have a combinatorial proof, like using inclusion-exclusion, but it seems not possible becuase some terms would not be integers. Would be glad if there is any proof or hint. Thanks.
 A: We have that for $0< j<n/2$,
$$\begin{align}
\sum_{k=0}^j (-1)^k \frac{n-2k}{n-j-k} \binom{n}{k} \binom{n-j-k}{j-k}&=
\frac{1}{n-2j} \sum_{k=0}^j (-1)^k (n-2k) \binom{n}{k} \binom{n-j-k-1}{j-k}\\
&=
\frac{n}{n-2j} \sum_{k=0}^j (-1)^k \binom{n}{k} \binom{n-j-k-1}{j-k}
\\&\qquad-\frac{2n}{n-2j} \sum_{k=1}^j (-1)^k \binom{n-1}{k-1} \binom{n-j-k-1}{j-k}.
\end{align}$$
Hence it suffices to show that
$$\sum_{k=0}^j (-1)^k \binom{n}{k} \binom{n-j-k-1}{j-k}
=2\sum_{k=1}^j (-1)^k \binom{n-1}{k-1} \binom{n-j-k-1}{j-k}\tag{$\star$}.$$
Now use the fact that
$$(-1)^k\binom{n-j-k-1}{j-k}=(-1)^j\binom{2j-n}{j-k}.$$
Then, by Vandermonde's identity, the LHS of $(\star)$  is
$$\sum_{k=0}^j (-1)^k \binom{n}{k} \binom{n-j-k-1}{j-k}=
(-1)^j\sum_{k=0}^j \binom{n}{k} \binom{2j-n}{j-k}=(-1)^j\binom{2j}{j}$$
and the RHS of $(\star)$ is
$$2\sum_{k=1}^j (-1)^k \binom{n-1}{k-1} \binom{n-j-k-1}{j-k}=
2(-1)^j\sum_{k=1}^j \binom{n-1}{k-1} \binom{2j-n}{j-k}
\\=2(-1)^j\binom{2j-1}{j-1}=(-1)^j\binom{2j}{j}$$
and we are done.
