Jtdylktuy

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.

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).

You can choose $4$ colors opposite the red and then $3$ colors opposite the blue. There are two colors left, which can be placed in $2$ ways onto the remaining two opposite fields. It follows that there are $4cdot3cdot 2=24$ different cubes that can be formed, given the two fields red and blue neighboring each other.

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.

1) That says nothing about the other colors; just that red is either above, below, to the left, or to the right, of the blue. 2) Because those are rotations THEY are redundant. You can count this as $1$ way. But no you have to figure out the different ways to put the remaining four colors.

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.

Huh????????

24: Similar concept to 4, just multiplied by 6.

Coincidentally that is the correct answer but for entirely the wrong meaning.

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Rotation doesn't matter. So you can always place the Blue face toward you. There are four faces adjacent to the Blue and one opposite. If the face opposite is Red that is one way to not do it. All other ways to paint the cube have the red face adjacent.

BUT because rotation doesn't matter those four faces are considered to be the same. So whichever face of those is red we can rotate it to the top.

But now we have rotated the cube so that the red face is on top. That we can not rotate it in any way and keep those two faces the in the same orientation.

There are four faces that need to be painted. Call them A,B,C,D. there are four choices of color for $A$. Once $A$ is chosen there are $3$ left for $B$ and so one. So there are $4*3*2*1 = 24$ to do this.

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It might be worth figuring how many ways in total there are to paint the cube.

Either the Red is adjacent to the blue (and there are $24$ ways to do that) or it is opposite the blue. If it opposite then the four remaining sides are all adjecent to both the blue and the red and the cube may be rotated so that any of them may be on top.

One of those four sides must be color 3. Paint one of them color 3 and rotate so color 3 is on top, Red is back, and Blue is facing you. The cube can not be rotated any further. There are $3$ faces left. Call them $A,B,C$. There are $3$ choices of color for face A, and after that $2$ for face $B$ and so on. So there are $3*2*1 = 6$ ways to have the Red face Opposite the Blue face.

So there are $24+6 = 30$ ways to pain the cube of which $frac 45$ of them have the Red adjacent to Blue.

The number of ways to reach a result depends on how many ways you are allowed to take,
that is you have to define the result (success) in a space of events.

In this case the wording suggests that you are considering the space of all the colored cubes
which are not equivalent upon rotation.

As explained in this related post such a space consists of $30$ different configurations.

In such space, you can always orientate the cube to have ,for instance, the R face in front of you.

Then there is only one way to get B non-adjacent to R: to be on the back face.

For the other colors, since you can rotate the cube around the RB axis, assume to fix the color of the upper face (e.g. yellow).

The choices you are left is how to place the other three colors, i.e. how to permute them, i.e. $6$ ways.

In fact, if we waive the adjacency requirement, we have $5$ ways to fix the color of the back face, since
a cube R front - B back cannot be rotated into a R front - not B back. For the lateral faces we are still given $6$ choices as above,
and so $30$ in total.

Thus $30-6=24$ are the ways to put R and B adjacent, which is your third result.

If you cut the cube open to get a configuration of squares as you say, then if you properly account for adjacencies and rotations
you shall still get $24$ out of $30$. If you get $10$ (don't know exactly under which hypotesis) something is wrong.