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In how many different ways can a cube be painted by using N different colors of paint?

Note that this question is not same to Painting the faces of a cube with distinct colours as the colours here may not be distinct (i.e. used many times). For Painting the faces of a cube with distinct colours, we must have $N>=6$, and the result will be:

$$\binom{n}{6}(5)(3!)$$

Can anyone help how to tackle such problem?

herohuyongtao
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2 Answers2

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See here: http://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem

Specifically, under Examples -> Colored cubes.

heropup
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I'm not sure if the answer is correct, but here's some of my idea: if we are painting a cube with $6$ distinct colours. fixing one of the faces, the face on the opposite side has $5$ colour choices using the circular arrangement for the rest of the $4$ sides we have $3!$ ways. In total, we have $5\times6=30$ ways.

Now if we have $n$ colours, after fixing the first face the opposite face has $(n-1)$ colour choices. There are now $(n-2)$ colours left to choose and for the four faces left we might choose $4$ colours from the $(n-2)$ colours and do the circular arrangements again.

In total, we have $(n-1)\times\binom{n-2}{4}\times3!$ ways.