What is the value of $ \sum\limits_{k=0}^{n-1}\binom {n-k-1}{j-1} \binom {r+k}{j+k}$? 
What is the value of  $$ \sum_{k=0}^{n-1}\binom {n-k-1}{j-1} \binom {r+k}{j+k}$$ where $r \ge j \ge 1$?

I know that 
$$ \sum_{k=0}^{n}\binom {n-k}{m}\binom{r+k}{s} = \binom {n+r+1}{m+s+1} \text{ where }n,m \ge0,\text{ and }s\ge r\ge 0$$
Please notice that second term of the first summation can be replaced by
$$ \binom {r+k}{r-j} $$
Then the first summation becomes
$$ \sum_{k=0}^{n-1}\binom {n-k-1}{j-1} \binom {r+k}{r-j} $$ which is identical to the second summation but doesn't satisfy the condition in second summation that is $ r \ngeq r - j$ since both $ r,j \ge 1 $
Is it possible to reduce the first summation similar to the second summation?

$$
\begin{align}
\sum_{k=0}^{n-1}\binom{n-k-1}{j-1}\binom{r+k}{j+k}
&=\sum_{k=0}^{n-1}\binom{n-k-1}{j-1}\binom{r+k}{r-j}
\end{align}
$$
which looks like $\binom{n+r}{r}$, but we are missing some $-j\le k\lt0$.

For example, let $n=6$, $j=3$, and $r=4$
$$
\begin{align}
\sum_{k=0}^{n-j}\binom{n-k-1}{j-1}\binom{r+k}{r-j}
&=\overbrace{\binom{5}{2}\binom{4}{1}}^{k=0}+\overbrace{\binom{4}{2}\binom{5}{1}}^{k=1}+\overbrace{\binom{3}{2}\binom{6}{1}}^{k=2}+\overbrace{\binom{2}{2}\binom{7}{1}}^{k=3}\\
&=95
\end{align}
$$
But $\binom{10}{4}=210$. Where is the missing $115$?
$$
\begin{align}
\sum_{k=-j}^{-1}\binom{n-k-1}{j-1}\binom{r+k}{r-j}
&=\overbrace{\binom{8}{2}\binom{1}{1}}^{k=-3}+\overbrace{\binom{7}{2}\binom{2}{1}}^{k=-2}+\overbrace{\binom{6}{2}\binom{3}{1}}^{k=-1}\\
&=115
\end{align}
$$
