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With three differently colored paints, in how many ways can the faces of a rectangular box can be painted so that the color changes occur only at each corner?

I was trying to solve this by principle of inclusion and exclusion, but I am unable to enumerate the number of ways to color the rectangular box( without any restrictions) because there are some distinct faces(adjacent faces) and some are non distinct faces(opposite faces).
Please help
Edit: The colorings which differ by rotation/s are considered to be same.

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    Exactly what do you mean when you say that *color changes occur only at each corner*?2012-12-10
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    Do you mean, *at each edge*?2012-12-10
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    Opposite faces could be distinct, I think. Are mirror-symmetric colorings considered identical?2012-12-10
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    Is there a rule that the colour *must* change where two walls meet? Also, I expect you can figure out the answer, the number is small. To simplify calculations, imagine that the West wall is painted red, and then multiply whatever answer you get by $3$.2012-12-10
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    @did yes it means edges2012-12-10
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    Six faces, three colors for each face, no other constraint... this does not seem so complicated. Or am I missing something?2012-12-10
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    Are you counting four walls or six? When you say *rectangular box* I think six, but some of the commentary suggests that you mean to exclude the top and bottom.2012-12-10
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    It says wall but surely that's too simple. But then again, ceiling and floor (??) is also too simple.2012-12-10
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    I was thinking it is like distribution of 6 object(denoting faces) in 3 distinct cell(denoting color) but again among those 6 objects some are distinct and some are not. Also opposite face could be distinct based on adjacent faces.2012-12-10
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    *So that color changes occur only at each corner* may simply mean that each of the four or six walls is painted a solid color, in which case the answer may be as simple as $3^4$ or $3^6$.2012-12-10
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    @BrianM.Scott doesn't $3^6$ counts some of cases more than once for example if east and west face is colored red and upward face is colored Blue then north and south face will be non distinct and (red,blue) , (blue,red) etc will be same but in $3^6$ ways it is counted as different. I think I am missing something silly can you help2012-12-10
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    If the six faces have individual identities (e.g., east, west, north, south, top, bottom), then no case is counted more than once in $3^6$. If, on the other hand, colorings that differ only by a rotation are supposed to be considered the same, then $3^6$ does overcount, badly. The real problem here is that I don’t really understand what the question is asking: there are too many possible interpretations, and I see no way to pick just one of them. If this is for a course, I think that you really need to get clarification from the instructor; if not, I don’t know what to suggest.2012-12-10
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    @BrianM.Scott This question is not from a course I just came across in some book. I was trying to solve this by taking the 2nd interpretation in which 2 coloring differ by rotation are same.2012-12-10
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    @anonymous no it doesn't have answers in the back :(2012-12-10

2 Answers 2

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Thought about it a bit more. There is precisely one way.

The formulation of the problem I am considering is that we have a rectangular box, three colors, say magenta, cyan, yellow, and we must color the sides of the box so that at every edge, the color of the walls adjacent to the edge is different.

In order to do this, we must have opposing sides be the same color. For suppose there are two opposing sides of different colors, say magenta and cyan. To be concrete, orient the box so that these sides are the "ceiling" and the "floor". Then consider the other four sides. Each of these is adjacent to a magenta side (the "ceiling") and a cyan side (the "floor"), so to avoid being the same color they must all be yellow. But then the sides are the same color as the other adjacent sides. So we must have opposing sides be the same color.

Now, from here, it is easy to see that any two cubes whose sides are colored like this can be rotated into the other. (Reflections don't come into account, since a cube colored like this is reflection-symmetric.) So pick any opposing pair of sides to be magenta, any other pair to be cyan, the last pair to be yellow; every choice is the same choice, so there is exactly one way to do this.

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    Sorry that I didn't stated the question clearly. I was taking the interpretation that 2 coloring differ by rotation are same.2012-12-10
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I assume you have a cube instead of an arbitrary box. You can use Burnside's lemma / Pólya's enumeration theorem. It's very nicely described here. Basically you find the different kind of rotations of the cube and count for each rotation how many colourings it fixes. Then Burnside's lemma states that the sum of these numbers divided by the number of rotations is the answer you seek.