You are allowed only to go east or north. Because of road construction, you cannot touch the points $a, b, c$ and $d$. Under these restrictions, the number of ways that you can go from $(0, 0)$ and finish at $(8, 8)$ in the following figure is:
First, I used ${16\choose 8}$ to get the number of ways without this construction. I am a bit confused on what to do next...
Through $a$. $\binom{6}{3}\times\binom{10}{5}$.
Through $b$ without going through $a$. $\binom{6}{4}\times\binom{9}{4}$.
Through $d$ without going through $a$. $\binom{6}{2}\times\binom{9}{5}$.
$\binom{16}{8}-\binom{6}{3}\times\binom{10}{5}-\binom{6}{4}\times\binom{9}{4}-\binom{6}{2}\times\binom{9}{5}=4050$
Here's how to solve this:
You are correct that there are ${16 \choose 8}$ paths without restrictions.
Now, consider paths that go through $a$. How many distinct initial path segments get you to $a$? There are ${6 \choose 3}$ of them, using the same logic of your broader solution. For each of these initial path segments, how many ways could you pass (ignoring the constraint)? There are ${10 \choose 5}$ of them (again by your logic). How many of these final paths go through $b$? Half of them! How many of these final paths go through $d$? The other half of them! These are all precluded. So now you know how many of the total number of paths are prevented, because they go through $a$.
Now, consider $d$. How many paths go through $d$ *but did not go through $a$? For that to happen, the path would have had to go through $(2,4)$, and then head east. (Do you see why?). Thus, use the same method to compute the number of paths that go through $(2,4)$ and then go east. For each of these, you can determine the number of unrestricted paths from $d$.
Do you get the basic logic here? Can you continue on your own?
