how do you answer :a bug lives on the corner of a cube and is allowed to travel only on the edges of the cube. in how many ways can the bug visit each of the other seven corners once and only once returning to its home corner only at the end of the trip?

Answer 1

I have not done an analytical solution, but just by intuition, I seem to get 12 as an answer. You can check if you can get more (or less).

Here's how I approach it.

Let the cube be represented by superimposing two squares, ABCD and EFGH where vertex A is above vertex E, and so on.

The bug lives at A and has to visit BCDEFGH and return to A in any order.

Let N=number of choices

For the first trip, it has three choices, AB,AD and AE. N=3 at this point.

For the route AB, on reaching B, it has two choices, BC and BF. By symmetry, it has now a total of N=3*2=6 routes.

If it chooses the route ABC, on reaching C, it has yet two more choices, CD and CG. So far, N=3*2*2=12 routes.

If it chooses ABCD, it will be forced to continue on to HGFEA if it wishes to visit the remaining vertices without repeating. If it chooses ABCG, he will have to continue with FEHDA for the same reason.

So the total number of routes it can take is 12.