This is an old paper by W.Feller.
It says that
$$ \sum_{k=r}^{n-1} \binom{k}{r}\binom{2n-2-k}{n-1} = \binom{2n-1}{n-1-r}$$
But I think, by Vandermonde's Identity, it's $\binom{2n-2}{n-1-r}$
Let $$p = k-r, then \sum_{p=0}^{n-1-r} \binom{k}{k-r}\binom{2n-2-k}{n-1-k} = \sum_{p=0}^{n-1-r} \binom{k}{p}\binom{2n-2-k}{(n-1-r)-p} = \binom{2n-2}{n-1-r}$$
Is it just a Typo or not?
$$ \begin{align} \sum_{k=r}^{n-1}\binom{k}{r}\binom{2n-2-k}{n-1} &=\sum_{k=r}^{n-1}(-1)^{k-r}\binom{-r-1}{k-r}(-1)^{n-1-k}\binom{-n}{n-1-k}\tag{1a}\\ &=(-1)^{n-r-1}\binom{-n-r-1}{n-r-1}\tag{1b}\\ &=\binom{2n-1}{n-r-1}\tag{1c} \end{align} $$ Explanation:
$\text{(1a):}$ negative binomial coefficients
$\text{(1b):}$ Vandermonde
$\text{(1c):}$ negative binomial coefficients
Vandermonde in the Upper Parameters
The Binomial Theorem says that $$ \begin{align} (1+x)^{-n-1}(1+x)^{-m-1} &=\sum_{j=0}^\infty(-1)^j\binom{n+j}{n}x^j\ \sum_{k=0}^\infty(-1)^k\binom{m+k}{m}x^k\tag{2a}\\ &=\sum_{j=0}^\infty\sum_{k=j}^\infty(-1)^k\binom{n+j}{n}\binom{m+k-j}{m}x^k\tag{2b}\\ &=\sum_{k=0}^\infty\sum_{j=0}^k(-1)^k\binom{n+j}{n}\binom{m+k-j}{m}x^k\tag{2c} \end{align} $$ and that $$ (1+x)^{-(n+m+1)-1}=\sum_{k=0}^\infty(-1)^k\binom{n+m+1+k}{n+m+1}x^k\tag3 $$ Putting $(2)$ and $(3)$ together, gives $$ \sum_{j=0}^k\binom{n+j}{n}\binom{m+k-j}{m}=\binom{n+m+k+1}{n+m+1}\tag4 $$ Thus, when the summation is carried out in the upper parameter of the binomial coefficients, an extra $1$ needs to be added in the upper and lower parameters in the resultant binomial coefficient, as compared to the standard Vandermonde Identity.