Consider a grid in $\mathbb{N}_0^2$. We can draw a path in it by traveling from point to point via a horizontal line segment to the right or vertical line segment going up. Let $k,n \in \mathbb{N}$ and such that $k \leq n/2$. I try to compute the number of paths from $(0,0)$ to $(n-k,k)$ that lie (non-strictly) under the diagonal and do not cross it. (That is, they may touch a point $(i,i)$, but not $(i,i+1)$.)
I have read in an article that the solution is ${n \choose k} - {n \choose k-1}$, but I have no idea how to prove this.
If we take a path from $(0,0)$ to $(n-k,k)$ which crosses the diagonal, it must at least touch the line $y=x+1$, so let P be the point where it first touches this line.
If we reflect the portion of the path from $(0,0)$ to P around the line $y=x+1$, we get a path from $(-1,1)$ to $(n-k,k)$.
Conversely, any path from $(-1,1)$ to $(n-k,k)$ must cross the line $y=x+1$, so if we let P be the point where the path first touches this line and reflect the portion of the path from $(-1,1)$ to P about this line, we get a path from $(0,0)$ to $(n-k,k)$ which crosses the diagonal.
Since this gives a bijection between the two types of paths, and since the number of paths from $(-1,1)$ to $(n-k,k)$ is given by $\binom{n}{k-1}$, we can conclude that the number of paths from $(0,0)$ to $(n-k,k)$ which do not cross the diagonal is given by $\binom{n}{k}-\binom{n}{k-1}$.