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Think about: If the cube is held in a particular orientation, there are 6! ways to paint the six faces. However, if you rotate the cube around, some of these colorings are equivalent. How many colorings are in each equivalence class?

So from what I understand, two colorings are equivalent provided that if the cube is rotated, they have the same coloring.

How can I tackle this problem? I know the strategy for anagrams, but maybe the context of this one is goofing me up. How can I find how many colorings are in each equivalence class?

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marked as duplicate by hardmath, Community Feb 24 '16 at 4:39

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  • $\begingroup$ I think, we can mark the faces as 1,2,3,4,5,6 and face 1 can be on up,down,front,back,left,right, there are 6 cases. Then when face 1 is fixed, there are 4 cases. So there are $6\times 4$ cases. The result might be $6!/24=30$, hope this might help $\endgroup$ – Alexis Feb 24 '16 at 4:16
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    $\begingroup$ Isn't this Related Question pretty much a duplicate? $\endgroup$ – hardmath Feb 24 '16 at 4:33
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pick one colour always for the top. eg black on top.

Now we have five choices for the bottom.

Now rotate so the darkest of the four remaining is on the left. We then have 3! ways to colour the last three. So I reckon

5 times 3! = 30

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