More help needed

Since there are only two ways to end to $B$, we have 2 cases.

Case 1: It ends with $\overline{FB}.$

There are 6 paths: $A \Rightarrow C \Rightarrow F \Rightarrow B$

$A \Rightarrow C \Rightarrow D \Rightarrow E \Rightarrow F \Rightarrow B$

$A \Rightarrow D \Rightarrow C \Rightarrow F \Rightarrow B$

$A \Rightarrow D \Rightarrow E \Rightarrow F \Rightarrow B$

$A \Rightarrow D \Rightarrow F \Rightarrow B$

$A \Rightarrow C \Rightarrow D \Rightarrow F \Rightarrow B.$

Case 2: It ends with $\overline{CB}.$

There are 4 paths: $A \Rightarrow D \Rightarrow C \Rightarrow B$

$A \Rightarrow D \Rightarrow F \Rightarrow C \Rightarrow B$

$A \Rightarrow C \Rightarrow B$

$A \Rightarrow D \Rightarrow E \Rightarrow F \Rightarrow B$

Adding these cases yields $6+4=\boxed{10 \text{ ways.}}$

If anyone finds a better way, please hit me up as this casework a bit tedious to do.

 - Jimmy