Consider an ant that is walking on a Cartesian grid, starting at (0, 0) and ending at (20, 12). The ant always chooses to walk exactly one unit either up or to the right (towards his destination) whenever he arrives at a Lattice point. (A Lattice point is a point with integer coordinates.) Thus, from (0, 0) he either walks to (1, 0) or (0, 1).
When doing a basic rectangle or a square like this, it's just the permutation of ups and rights total, so I found it to be $$\frac{32!}{20!12!}$$ Heres where I'm getting a little confused.
If the ant is not allowed to go to the points (10, 5) and (12, 8), how many different paths can he take on his walk?
When asked to exclude certain points, do I have to subtract the total number of ways to get to each of these points from the total amount of ways to get to the end? Would that cause an overlap that I would have to add back in? How do I know when doing inclusion and exclusion problems when to add back?
Let $T$ be the total number of paths without removing anything.
Removing a point $A$ removes all paths which pass through $A$ (nevermind any other second removal for now). Say $T_A$ is this value.
Similarly define $T_B$.
If you take $T - T_A - T_B$, then you have doubly-removed all paths which pass through both $A$ and $B$. So you need to add back the number of paths which pass through both $A$ and $B$, say $T_{AB}$. The answer you're looking for is $T - (T_A + T_B) + T_{AB}$.