Combinatorial proof that $(n-r){n+r-1 \choose r}{n \choose r} = n{n+r-1 \choose 2r}{2r \choose r}$.
Typically to combinatorially prove something we need to show that the LHS indeed counts the same thing as the RHS. I am having trouble doing so. One thing I noticed is that on the LHS, we are choosing from a total of $(n-r)+(n+r-1)+(n) = 3n-1$, but on the right we chose from $2n+3r-1$. This is likely no problem at all, but it is still roadblock for me.
Any tips or hints on how to approach this problem are appreciated