I am trying to show $$f(n,n)=\sum_{i=0}^{n-1}{n-1\choose i}{n+i-1\choose i}$$ for $f(n,n)$ the central delannoy number. I am stuck at the following. We have $${2n-k\choose k,n-k,n-k}=\frac{(2n-k)!}{(n-k)!(n-k)!k!}=\frac{(2n-k)!n!}{(n-k)!(n-k)!k! n!}={2n-k\choose n}{n\choose k}$$ the number of ways to get to $(n,n)$ with a fixed number of diagonal steps $k$.
Then we sum over $k$ or $i$ $$f(n,n)=\sum_{i=0}^{n}{2n-i\choose n}{n\choose i}$$ not sure where to go from here
The second sum is \begin{eqnarray*} \sum_{k=0}^{n} \binom{2n-k}{n-k} \binom{n}{n-k}. \end{eqnarray*} Reverse the order of this sum with the substitution $i=n-k$ and we have \begin{eqnarray*} \sum_{i=0}^{n} \binom{n+i}{i} \binom{n}{i}. \end{eqnarray*} (Check the first formula needs $n-1 \rightarrow n $) https://en.wikipedia.org/wiki/Delannoy_number#Central_Delannoy_numbers_2