Number of paths in a grid below a diagonal [closed]

We know that there are $\binom{2n}{n}$ ways of going $(n,n)$ from $(0,0)$ when there is no restriction. So the answer should be $\binom{2n}{n}-B$ where $B$ is the number of "bad paths", that is, number of paths that go above the diagonal line.

Now, in order to calculate $B$, we should notice something: When we go above the diagonal line, we will pass through another diagonal line, which is shown as $\color{red}{red}$ in the figure. And for each bad path (passes through the red line), let us create a new path which has the same moves until we go above the diagonal line (or pass through the red line), and then does the symmetric moves to our original path (original path is shown as $\color{blue}{blue}$, symmetric moves are shown in $\color{lime}{green}$). Therefore, number of bad paths become the number of paths from $(0,0)$ to $(n-1,n+1)$, which means $B = \binom{2n}{n+1}$. So the answer becomes $$\binom{2n}{n}-\binom{2n}{n-1}$$ which is also called the $n^{th}$ Catalan Number.