This problem can be solved using the Fundamental Principle of Counting or FPC. The FPC states that if one event can be done in [tex]m[/tex] ways and another event can be done in [tex]n[/tex] ways, the number of ways that the two events can happen is [tex]mn[/tex] ways. For example, if you have 2 different shirts and 5 different pair of pants, the number of possible combinations of a shirt and a pair pants that you can make is 2×5 = 10 ways.
Now, to solve this problem, we have to remember that the route is this: From A to B, then from B to C, then from C back to B, and from B back to A.
Let's count our choices:
(a) From A to B, we have 3 roads to choose from;
(b) From B to C, we have 4 roads to choose from;
(c) In going back from C to B, we only have 3 options this time since we already took one in (b), again we cannot take the same road twice;
(d) From B to A, we only have 2 roads to choose from since we already took one in (a).
Therefore, applying the FPC now, we have a total of 3×4×3×2 = 72 ways.
See another problem that uses FPC: brainly.ph/question/279814
