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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 $3 \times 3$.question involving graph, need help on part c

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.

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  • $\begingroup$ 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! $\endgroup$
    – Don Larynx
    Nov 2 '13 at 21:53
  • $\begingroup$ @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. $\endgroup$
    – Celeritas
    Nov 2 '13 at 21:58
  • $\begingroup$ $R_5$ is one path, what are you on mate? $\endgroup$
    – Don Larynx
    Nov 2 '13 at 22:00
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    $\begingroup$ Crack cocaine.. $\endgroup$
    – Celeritas
    Nov 2 '13 at 22:09
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    $\begingroup$ 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.] $\endgroup$
    – coffeemath
    Nov 3 '13 at 1:35
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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.

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  • $\begingroup$ The second part of the question mentions "to calculate the number of different round trips" won't the answer for that part be 14*13 = 182, I say this because the question doesn't mention to consider repetition of paths. $\endgroup$ Jul 4 '14 at 18:57
  • $\begingroup$ @AnkitSablok No: Different round trips is 14*14, that is, one of 14 ways A to C followed by one of 14 ways from C to A. By saying "different round trip" means the overall trips are different. In part (c) the extra requirement that the return trip differs from being the exact reverse of the initial trip thus rules only one of them out. $\endgroup$
    – coffeemath
    Jul 4 '14 at 23:23
  • $\begingroup$ You are correct in saying that, but the language of the question seemed a little ambiguous to me :) $\endgroup$ Jul 5 '14 at 0:56

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