Lattice paths with avoiding a group of points My house is located at $(0,0)$ and the destination is located at $(n, n)$ for $m, n \in \mathbb{Z}^+$. I can only move in the upwards or in rightward directions. How many paths are there if I want to avoid the really busy group of points located at $(x, y)$ such that for $0 \leq x < m, n-m < y \leq n$.
This question seems very similar to the Dyck paths problem for avoid lines over a diagonal $y=x$ where we need to find the total paths that are $\binom{2n}{n}$ and we need to subtract the bad paths. It seems like the bad paths are in a square-like grid so the number of bad paths should be: $$\binom{n}{m}$$
Please correct me if I am wrong and if you could explain for a concrete example like $m=3, n=6$ that would be helpful.
 A: If I understand the problem correctly, the points to be avoided form a square in the upper lefthand corner, specifically, the square
$$\{0,1,\ldots,m-1\}\times\{n-m+1,n-m+2,\ldots,n\}\;.$$
For $n=6$ and $m=3$ we’d have the lattice points shown below; the block to be avoided is shown in red.
$$\begin{array}{c|cc}
6&\color{red}\bullet&\color{red}\bullet&\color{red}\bullet&\bullet&\bullet&\bullet&\bullet\\
5&\color{red}\bullet&\color{red}\bullet&\color{red}\bullet&\bullet&\bullet&\bullet&\bullet\\
4&\color{red}\bullet&\color{red}\bullet&\color{red}\bullet&\bullet&\bullet&\bullet&\bullet\\
3&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet\\
2&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet\\
1&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet\\
0&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet\\\hline
&0&1&2&3&4&5&6&
\end{array}$$
Every bad path must hit the bottom row of bad points, i.e., one of the points $\langle k,n-m+1\rangle$ with $0\le k<m$. It might hit more than one of those points, since it can move to the right after hitting that row, but in any case there must be a first point $\langle k,n-m+1\rangle$ at which it hits that row. And since that’s the first point, the path must hit it from below, i.e., from the point $\langle k,n-m\rangle$. Thus, we can classify the bad paths by where they move up from the line $y=n-m$ to the line $y=n-m+1$.
For $0\le k<m$ there are $\binom{k+n-m}{n-m}$ paths from $\langle 0,0\rangle$ to $\langle k,n-m\rangle$, so there are $\binom{k+n-m}{n-m}$ bad paths that enter the prohibited block from $\langle k,n-m\rangle$. Each of those paths then goes to $\langle k,n-m+1\rangle$, from which point it can reach $\langle n,n\rangle$ in
$$\begin{align*}
\binom{(n-k)+(n-(n-m+1))}{n-k}&=\binom{n+m-1-k}{n-k}\\
&=\binom{n+m-1-k}{m-1}
\end{align*}$$
different ways. Thus, there are
$$\sum_{k=0}^{m-1}\binom{n-m+k}{n-m}\binom{n+m-1-k}{m-1}$$
bad paths. Unfortunately, I don’t immediately see a closed form for this sum.
