King on chessboard part 2 In continuation to King on chessboard question
Short summary of first part - we have chessboard and king on a1. King can ONLY move one space up, or one space right. There is exactly $\binom{7+7}{7}=3432$ ways king can arrive at h8
Setting is exactly the same, but with quirk that our hypotetical king cannot move onto some square in the middle of the board, lets say - king cannot step on d6.
My idea was to count all possible routes to c5, then to have two choices: twice right or twice up, and then count all routes from c7 and e5 to h8.
Then it occured to me, that king doesnt NECESSARILY has to go through c5 at all. Ie RRRRRRRUUUUUUU is route which is absolutely valid, yet uncounted in my thoughts.
What am I missing here?
 A: You know there are a total of $\binom{14}{7}$ paths so by complementary counting you need to find those who pass through d6  and then subtract them from the total number .
Let's take a path that passes through d6 .Now break the path into two parts :
$1)$ The part from a1 to d6
$2)$ The part from d6 to h8
Now just as in the original problem :


*

*To land on d6 you must make three steps up and five steps right so there are $\binom{3+5}{3}$ ways to form the first part of the path.

*To continue from d6 to h8 you must make four steps up and two steps right so there are $\binom{4+2}{2}$ ways to form the second part of the path .
This means there are : $$\binom{8}{3} \cdot \binom{6}{2}$$ paths that pass through d6 and thus there are  :
$$\binom{14}{7}-\binom{8}{3} \cdot \binom{6}{2}$$ paths that don't pass through d6 . 
A: Use your previous technique to count the paths that go through d6, then subtract them from your previous result. To go through d6, you first go from a1 to d6, then from d6 to h8
