I'm trying to see if I'm correct with a general approach to solving these lattice path problems using the commonly known reflection proof and the Catalan numbers.
If we want to see how many paths we can take while touching the diagonal line $y=x$ it's $C_n = \frac{1}{n+1}\binom{2n}{n}$.
If we want to consider all the paths that do not touch the line except for the starting position of $(0,0)$ and end position $(n,n)$. Is that the same as considering all the paths that touch the line $y = x + 1$? Can we use a similar method as that proof to answer a similar question where the starting point is at $y=x$, but we cant touch the line $y = x + 2$ or $y = x + 3$? Where the construct the bad paths just by shifting the n and m appropriately?
I have this picture drawn to exemplify what I'm talking about...
