smc

Note that cities $C$ and $E$ can be removed when counting paths because if a path goes in to $C$ or $E$, there is only one possible path to take out of cities $C$ or $E$. So the diagram is as follows: Now we proceed with casework. Remember that there are two ways to travel from $A$ to $D$, $D$ to $A$, $B$ to $D$ and $D$ to $B$.: Case 1 $A\Rightarrow D$: From $D$, if the path returns to $A$, then the next path must go to $B\Rightarrow D\Rightarrow B$. There are $2\cdot 1\cdot 2=4$ possibilities of the path $ADABDB$. If the path goes to $D$ from $B$, then the path must continue with either $BDAB$ or $BADB$. There are $2\cdot 2\cdot 2=8$ possibilities. So, this case gives $4+8=12$ different possibilities. Case 2 $A\Rightarrow B$: The path must continue with $BDADB$. There are $2\cdot 2=4$ possibilities for this case. Putting the two cases together gives $12+4=\boxed{16}$