We have a dude who throws a dice at every time he reaches an intersection, and he is starting at the bottom left (if you look closely you can see 'start'). If he throws a 6, he'll move in an eastward direction and if he throws less than 6 he'll move in a northward direction.
You might see this coming, what are the chances that he reaches $A$, $B$ and $C$ after 8 throws? We'll only discuss $A$.
My textbook says the answer is $(\dfrac{1}{6})^2 \times (\dfrac{5}{6})^4 \times \binom{8}{2}$
What I don't understand is the $\binom{8}{2}$ part. We arrive at $A$ when we have $(\dfrac{1}{6})^2 \times (\dfrac{5}{6})^4$, so when add 2 moves, isn't there a (big) possibility that we arrive somewhere else?