Mr. P has decided to visit the problem solvers from the following cities because they have sent in the most correct answers in the past year:
Flint, Michigan; Pune, India; Delta, Colorado; Mountain View, Wyoming; Mumbai City, India; Joilet, Illinois; and Houston, Texas.

Mr. P only wants to use the following direct routes between the cities as shown in the map below:



Here are the eight cities showing the routes but not drawn to scale:



Mr. P must start and stop in Fort Collins, Colorado, and he may pass through a city more than once.

In how many ways can he visit all the cities without retracing any route?

List the ways in which he can do this.   You may use the following abbreviations:
F, P, D, MV, MC, J, H, and FC.

In lieu of listing all the ways, you may provide a detailed explanation.
Solution to the Problem:

The answer is 48 ways:

FC-MV-D-FC-F-J-H-MC-P-H-FC
FC-MV-D-FC-F-J-H-P-MC-H-FC
FC-MV-D-FC-F-H-MC-P-H-J-FC
FC-MV-D-FC-F-H-P-MC-H-J-FC

FC-MV-D-FC-J-F-H-MC-P-H-FC
FC-MV-D-FC-J-F-H-P-MC-H-FC
FC-MV-D-FC-J-H-MC-P-H-F-FC
FC-MV-D-FC-J-H-P-MC-H-F-FC

FC-MV-D-FC-H-P-MC-H-J-F-FC
FC-MV-D-FC-H-P-MC-H-F-J-FC
FC-MV-D-FC-H-MC-P-H-J-F-FC
FC-MV-D-FC-H-MC-P-H-F-J-FC

FC-D-MV-FC-F-J-H-MC-P-H-FC
FC-D-MV-FC-F-J-H-P-MC-H-FC
FC-D-MV-FC-F-H-MC-P-H-J-FC
FC-D-MV-FC-F-H-P-MC-H-J-FC

FC-D-MV-FC-J-F-H-MC-P-H-FC
FC-D-MV-FC-J-F-H-P-MC-H-FC
FC-D-MV-FC-J-H-MC-P-H-F-FC
FC-D-MV-FC-J-H-P-MC-H-F-FC

FC-D-MV-FC-H-P-MC-H-J-F-FC
FC-D-MV-FC-H-P-MC-H-F-J-FC
FC-D-MV-FC-H-MC-P-H-J-F-FC
FC-D-MV-FC-H-MC-P-H-F-J-FC

FC-F-J-H-MC-P-H-FC-D-MV-FC
FC-F-J-H-P-MC-H-FC-D-MV-FC
FC-F-J-H-MC-P-H-FC-MV-D-FC
FC-F-J-H-P-MC-H-FC-MV-D-FC

FC-F-H-MC-P-H-J-FC-MV-D-FC
FC-F-H-MC-P-H-J-FC-D-MV-FC
FC-F-H-P-MC-H-J-FC-MV-D-FC
FC-F-H-P-MC-H-J-FC-D-MV-FC

FC-J-F-H-MC-P-H-FC-D-MV-FC
FC-J-F-H-P-MC-H-FC-D-MV-FC
FC-J-F-H-MC-P-H-FC-MV-D-FC
FC-J-F-H-P-MC-H-FC-MV-D-FC

FC-J-H-MC-P-H-F-FC-MV-D-FC
FC-J-H-MC-P-H-F-FC-D-MV-FC
FC-J-H-P-MC-H-F-FC-MV-D-FC
FC-J-H-P-MC-H-F-FC-D-MV-FC

FC-H-MC-P-H-F-J-FC-MV-D-FC
FC-H-MC-P-H-F-J-FC-D-MV-FC
FC-H-MC-P-H-J-F-FC-MV-D-FC
FC-H-MC-P-H-J-F-FC-D-MV-FC

FC-H-P-MC-H-F-J-FC-MV-D-FC
FC-H-P-MC-H-F-J-FC-D-MV-FC
FC-H-P-MC-H-J-F-FC-MV-D-FC
FC-H-P-MC-H-J-F-FC-D-MV-FC

Many thanks to Rob Miles for correcting my original answer!

Here is Rob's excellent analysis of the problem:

The first way I got to this answer was by breaking the figure into three sections. (See picture.) Section A is MV, D, FC. Section B is the middle of the figure, including J, F and the paths that connect FC to H. Section C is H, MC, P.



Sections A and C are both loop paths. You have to get to each of those cities, so you have to go around the loop. You can either go around the loop clockwise or counterclockwise. So both of these sections can be traversed in 2 ways.

Section B is the trickiest. You have to go through section B to get to section C, then back through section B to get home. While going through section B on either the way out or the way back, you have to get to both J and F. There are six ways to do this. 

Count   Out   Back
1       J,F    -
2       F,J    -
3        J     F
4        F     J
5        -    J,F
6        -    F,J
Considering the entire figure at once, there are only two allowable orders to traverse the sections. You either go ABCB or you go BCBA.

To get an answer, we have to multiply all of those options together: the allowable orders of the sections times each of the allowable ways to traverse each section.

2 x 2 x 6 x 2 = 48 ways

I was able to confirm this answer using a different approach. This one was a bit more intricate. I have already written quite a bit here, so I will only summarize this. You can figure out the details if you are curious enough to do so. There are 12 paths drawn on your figure, however all of the allowable solutions will use exactly 10 of those paths. There are exactly 3 pairs of paths that could be left out. You could leave out (1) FC-J and F-H or (2) FC-F and J-H or (3) FC-H and J-F. In each of these 3 cases, you would now have a figure where most of the cities only have two paths. If a city has only two paths, it is trivial - you enter through one path and exit through the other path. If a city has more than two paths, it needs to be considered. With any of the path pairs removed, FC and H are the only cities with more than two paths remaining. The first time you are in FC you have 4 options; the second time you are in FC you have 2 options; the first time you are in H you have 3 options, but 1 of them would be incorrect, so only 2 options; the second time you are in H you have only 1 option; the third time you are in FC the puzzle is completed. Finally, 3 ways to remove path pairs times 4 options in FC times 2 options in FC times 2 options in H: 3 x 4 x 2 x 2 = 48!

Correctly solved by:

1. Rob Miles Northbrook, Illinois
2. Ivy Joseph Pune, Maharashtra, India