Is this combinatorics question correct? 
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?
 A: That calculation is actually a mixture of the calculations for $A$ and $B$. In order to get from the starting point to $B$, you must take exactly $2$ steps to the east and $6$ to the north. It doesn’t matter in what order you take those $8$ steps: they will always get you from $\text{start}$ to $B$.
There are $\binom82$ different $2$-element subsets of any $8$-element set, so there are $\binom82$ ways to choose which $2$ of your $8$ steps will go to the east; by default the other $6$ steps will be to the north. However, the probability of any given sequence of $2$ eastward steps and $6$ northward steps is $\left(\frac16\right)^2\left(\frac56\right)^6$, not $\left(\frac16\right)^2\left(\frac56\right)^4$, and the total probability of getting to $B$ is therefore
$$\left(\frac16\right)^2\left(\frac56\right)^6\binom82\;.$$
To get to $A$, on the other hand, you must take one step to the east and four to the north. There are $\binom51=5$ ways to place the eastward step in the string of $5$ steps: it can be first, second, third, fourth, or last. The probability of any specific string of $5$ steps, one to the east and four to the north, is $\left(\frac16\right)^1\left(\frac56\right)^4$, so the total probability of reaching $A$ is
$$\left(\frac16\right)^1\left(\frac56\right)^4\binom51\;.$$
