Number of paths in a board with a puncture A picture of the board and my solution

I have this $8 \times 6$ board but there is a restriction with it - we have a hole with a bottom-left corner at $(3, 2)$ and top-right at $(6, 4)$. I need to count the number of possible ways to get from the top-left corner to the bottom-right corner of the board.
Since there are two ways to go around the hole, I counted them and summed them. Is my solution correct or did I miss something?
Thank you in advance!
 A: The usual trick here is to mark diagonal lines of cells such that all paths pass through exactly one of those cells.
S.......1
.......2.
......3..
....  ...
...4.....
..5......
.6......E

Now the result is a sum of products of binomials:


*

*$\binom80\binom60=1$ path goes through cell 1.

*$\binom81\binom61=48$ paths go through cell 2.

*$\binom82\binom62=420$ paths go through cell 3.

*$\binom73\binom72=735$ paths go through cell 4.

*$\binom72\binom71=147$ paths go through cell 5.

*$\binom71\binom70=7$ paths go through cell 6.


Thus there are 1358 paths in total.
A: 
Here is a variant which evaluates  the number of valid lattice paths by iteratively adding up the number of valid paths with  shorter length.
                   
We conclude  in concordance with the  answer of @ParclyTaxel:
The number of valid paths is $$\color{blue}{1358}$$

A: If the board did not contain a hole, any path would involve eight moves to the right and six downward moves.  The path is completely determined by choosing which six of the fourteen moves are downward.  Hence, without the hole, there would be $$\binom{14}{6}$$ possible paths. 
From these, we must exclude those paths that pass through the two missing vertices.  With your labeling system, these vertices are at $(4, 3)$ and $(5, 3)$.  
Paths that would pass through the vertex $(4, 3)$:  The vertex $(4, 3)$ is four units to the right and three units down from the top left corner, so there would be $\binom{7}{3}$ ways to reach the vertex from the left corner.  The vertex $(4, 3)$ is also four units to the left and three units up from the bottom right corner, so there would be $\binom{7}{3}$ ways to reach the bottom right corner from $(4, 3)$.  Hence, the number of paths that would pass through the vertex $(4, 3)$ would be 
$$\binom{7}{3}\binom{7}{3}$$  
Paths that would pass through the vertex $(5, 3)$:  The vertex $(5, 3)$ is five units to the right and three units from the top left corner, so there would be $\binom{8}{3}$ ways to reach the vertex from the top left corner.  The vertex $(5, 3)$ is three units to the left and three units up from the bottom right corner, so there would be $\binom{6}{3}$ ways to reach the bottom right corner from $(5, 3)$.  Hence, the number of paths that would pass through the vertex $(5, 3)$ would be 
$$\binom{8}{3}\binom{5}{3}$$
However, if we simply subtract the number of paths that pass through these vertices from the total, we will have subtracted paths that pass through both of these vertices twice.  We only want to subtract those paths once, so we must add them back.
Paths that would pass through both the vertices $(4, 3)$ and $(5, 3)$:  There are $\binom{7}{3}$ ways to reach $(4, 3)$ from the upper left corner, one way to reach $(5, 3)$ from $(4, 3)$, and $\binom{6}{3}$ ways to reach the bottom right corner from $(5, 3)$.  
Hence, by the Inclusion-Exclusion Principle, the number of permissible paths is 
$$\binom{14}{6}  - \binom{7}{3}\binom{7}{3} - \binom{8}{3}\binom{6}{3} + \binom{7}{3}\binom{6}{3}$$
