Each face of a cube can be painted in one of six colors and the color of each face must be different. Suppose you pick a coloring uniformly at random from the set of allowed coloring. How many ways can the cube be arranged such that the red face is adjacent to the blue face, assuming red and blue are among the six allowed colors.(Keeping rotation in mind so A=red B=blue is the same as A=blue, B=red, and counts as one case)
I am getting confused, with 4, 10 and 24 as possible answers.
4: Since the cube can be rotated, assume B is fixed to the front face of the cube. Hence, there are 4 possible ways R can be oriented around B. Since there are rotations, it is redundant to multiply by the 6 faces on the cube. Hence 4 ways.
10: If I map out the cube by drawing 4 horizontal boxes and 1 box below and 1 box above, and assign B to each face, I count the number of ways R can be next to B for each box, and arrive at 10.
24: Similar concept to 4, just multiplied by 6.
I am really confusing myself here and it would really help if someone could clarify this for me as I am really bad at spatial visualisation.
Best Answer
If the red and the blue are next to each other, then it's possible to orient the cube so that the red is on top and the blue is on the front face. From this, all $4! = 24$ arrangements of the remaining four colors give a unique coloring of the cube (down to rotation of the cube).