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Consider a dodecahedron with its faces numbered from 1 to 12. Each of those faces is going to be painted, having for such 8 different colors available.

How many different ways is it possible to color the dodecahedron, supposing that 6 of its faces must be painted the same color and the remaining ones must have different colors, either between them and the 6 that share the same color?

I did $$^{12}C_6*^7P_6$$

However, my book says the solution is $$8 * ^{12}C_6*^7P_6$$

Why do I have to multiply by 8? I don't understand.

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Segmentation fault
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  • I am not sure what ${}^{7}!!A_6$ represents, by you could have $8$ for the main colour, $7$ for the unused colour, ${}^{12}!C_6$ for the faces with the main colour and $6!$ for the pattern/order of the other colours on the other six faces – Henry Oct 14 '16 at 17:10
  • I'm not used to that notation, but basically for each color of the 8 avaliable, you count every subset of faces. For each pair of color and subset, you can pick 6 out of the 7 remaining colors, and paint the 6 remaining faces of 6! different ways. – qualcuno Oct 14 '16 at 17:11
  • @Henry The A would be $P$... the book uses different notation. I'll change it – Segmentation fault Oct 14 '16 at 17:15
  • @SilenceOnTheWire OK, so $8$ ways of choosing the colour for the six faces coloured the same – Henry Oct 14 '16 at 17:17
  • @Henry yes. But why did you use $6!$ for the other colors? If you have 8 colors and use one, you still have 76543*2 possible permutations, which is not $6!$... – Segmentation fault Oct 14 '16 at 17:24
  • @SilenceOnTheWire: I took $7$ for the unused colour and then $6!$ for the pattern. In effect you had $\frac{7!}{1!}$ for the pattern, so again $7!$ but expressed a different way. – Henry Oct 14 '16 at 17:33
  • @Henry Ah! I see – Segmentation fault Oct 14 '16 at 17:37
  • See the general answer given by Marko Riedl in (http://math.stackexchange.com/q/53859) and take $n=8$. – Jean Marie Oct 14 '16 at 22:09

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