I'm attempting the following question:
How many paths are there where you can only move up or right one unit at each step that go from (0, 0) to (5, 3) without crossing the line y = x? You are allowed to touch it.
Without considering y=x, we would have 56 ways to reach (5,3). How do I frame this question to approach an answer?
Let $n=4$ and solve the general case of $(n+1, n-1)$ instead of $(5,3)$.
As in the catalan proof reflect paths across $y=x-1$.
You will get exactly all paths from $(0,0)$ to $(n,n)$ that begin with a right step, except for those who exceed the reflection of $y=x$ which is $y=x-2$ i.e those who intersect with $y=x-3$. Call these paths bad.
There are ${2n-1 \choose n}$ non-bad paths.
Reflect accrose $y=x-3$ and you will see every bad path correspond to a normal path starting with a right step from $(0,0)$ to $(n+3,n-3)$. Hence there are ${2n-1 \choose n+3}$ bad paths.
Overall ${2n-1 \choose n}-{2n-1 \choose n+3}$ which is $28$ in case $n=4$ $\blacksquare$
