Number of possible paths on a lattice grid given a restriction 
On a $6 \cdot 6$ lattice grid, an ant is at point $(0,0).$ There is a teleportation pad at $(2,2)$ and $(3,3).$ When an ant reaches either teleportation pad, it teleports the ant from the pad it is on to the other pad, and then the pad disappears. (Meaning the pad can only be used once) For example, if the ant reaches $(2,2),$ it is automatically brought to $(3,3).$ How many possible routes are there to take from $(0,0)$ to $(5,5),$ such that the ant can only move up or rightwards along the lattice grid.

There is a formula for finding the possible routes from one point to another without having teleportation. I considered the problem if the teleportation condition was voided. Then, the problem would simply become $\binom{10}{5}=252.$ However, I feel like I have overcounted some cases and am not sure how to subtract them. May I have some help? Thanks in advance.
 A: *

*How many paths are there that avoid both teleportation pads?

*How many paths are there from $(0,0)$ to the first pad at $(2,2)$?

*How many paths are there starting from the second pad at $(3,3)$ to $(5,5)$?  (Hint:  compare with #2 above.)

*How many paths are there from $(0,0)$ to the second pad at $(3,3)$, avoiding $(2,2)$?

*How many paths are there starting from $(2,2)$ to $(5,5)$, possibly through $(3,3)$?


To get from $(0,0)$ to $(x,y)$, there are $$\binom{x+y}{x} = \frac{(x+y)!}{x! y!}$$ unrestricted possibilities.
There are $\binom{4}{2} = 6$ paths from $(0,0)$ to $(2,2)$, and an equal number of paths from $(3,3)$ to $(5,5)$.  And there are $\binom{2}{1} = 2$ paths from $(2,2)$ to $(3,3)$.  So there are $\binom{6}{3} - \binom{4}{2}\binom{2}{1} = 20 - 12 = 8$ paths from $(0,0)$ to $(3,3)$ that do not pass through $(2,2)$, and the same number from $(2,2)$ to $(5,5)$ that do not pass through $(3,3)$.
Consequently, ignoring teleportation, there are $8 \binom{4}{2} = 48$ paths from $(0,0)$ to $(5,5)$ that pass through $(3,3)$ do not pass through $(2,2)$, and an equal number of paths that pass through $(2,2)$ but not $(3,3)$.  There are $\binom{4}{2}\binom{2}{1}\binom{4}{2} = 72$ paths that pass through both $(2,2)$ and $(3,3)$.  All combined, there are $\binom{10}{5} - 2(48) - 72 = \color{red}{84}$ valid paths that do not pass through either $(2,2)$ or $(3,3)$.
Second, count the number of paths to $(5,5)$ through $(2,2)$ and teleporting to $(3,3)$.  This is just $\binom{4}{2} \binom{4}{2} = \color{red}{36}$.
Third, we know that the answer to item number 4 above is just $8$.  After teleportation back to $(2,2)$, there are another $\binom{6}{3}$ ways to get to $(5,5)$, noting that after teleportation, $(3,3)$ becomes an ordinary point.  So this last case has $8(20) = 160$ paths.
All put together, there are $84 + 36 + 160 = \color{red}{280}$ paths.
