Monotonic Lattice Paths and Catalan numbers Can someone give me a cleaner and better explained proof that the number of monotonic paths in an $n\times n$ lattice is given by ${2n\choose n} - {2n\choose n+1}$ than Wikipedia
I do not understand the how they get ${2n\choose n+1}$ and I do not see how this is the number of monotonic paths that cross the diagonal. Please be explicit about the $n+1$ term. I think this is the hardest part for me to understand.
I understand the bijections between proper parenthesizations and so forth...
Thanks!

Consider the northeastern lattice paths from $(0, 0)$ to $(n, n)$ which do cross the diagonal. In any of these paths there must be a first edge (say, the $i^{th}$ edge) which crosses the diagonal. Now consider the portion of the path that occurs after the $i^{th}$ edge, and reflect this portion in the axis $y = x + 1$. The resulting (full) path is a northeastern lattice path from (0, 0) to (n−1,n+1).
Since every monotonic path from $(0, 0)$ to $(n − 1, n + 1)$ must cross the diagonal at some point, this reflection gives us a bijection between northeastern lattice paths from $(0,0)$ to $(n, n)$ which cross the diagonal and northeastern lattice paths from $(0, 0)$ to $(n − 1, n + 1)$. There are $\binom{2n}{n+1}$ such paths. Therefore, the number of paths never cross the diagonal is $\binom{2n}{n}-\binom{2n}{n+1}$.
