# How many ways is there to do a round trip if at least one of the roads taken on the return trip is different?

I'm stuck on part c) of this question. The answer key gives 182. I already know there are 14 ways to make the trip from city A and city B and vice versa. It appears 182 came from $14 \times 13$ but I don't get where 13 came from? If neither $R_8$ or $R_9$ were used on the trip there, one of the roads in the middle would need to be removed from the return trip so that would become $2 \times 4$ or $3 \times 3$.

By the way, what's the policy for this site regarding typing out questions instead of putting images for them? Normally I would try but if a person can't view images then they wouldn't be able to see the graph and wouldn't be able to help anyways.

• Hint: Round trip possibilities: #Trip forward $\times$ #trip backward. Thus 14 * 14-1 (1 road is different, mainly the original 1). Images are great if they are clear. As long as you show your work, or at an attempt at minimum! Nov 2 '13 at 21:53
• @DonLarynx but I disagree; it shouldn't be 14-1 because if you take off R5 fore example, then there isn't 13 routes but 10. Nov 2 '13 at 21:58
• $R_5$ is one path, what are you on mate? Nov 2 '13 at 22:00
• Crack cocaine.. Nov 2 '13 at 22:09
• If there are $14$ ways from $A$ to $C$, to count the round trips not using exactly the same route backwards from $C$ to $A$, do it by first choosing one of the $14$ ways from $A$ to $C$, and then choosing one of the $13$ ways which differ from the way you already took from $A$ to $C$, and going back from $C$ to $A$ by that route. There are then $14*13$ ways as the answer says. [There is no need to get into specifics about the form of the routes involved, each route is reversible.] Nov 3 '13 at 1:35

To plan a round trip from $A$ to $C$ and back, where the return route is not the reverse of the first part from $A$ to $C$, we proceed as follows. Since there are $14$ ways to go from $A$ to $C$ we must choose one of those ways. Once that has been done, exactly one of the $14$ routes has been ruled out for reversal to get back to $A$ from $C$, leaving us $13$ choices for the return trip. This gives $14\cdot 13=182$ possible round trips.
Note that we do not have to consider the specific types of these trips in terms of how they go through $B$ (or avoid $B$). One of the trips for the first leg of the journey from $A$ to $C$ being chosen, it is the only one excluded on the way back.