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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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  • $\begingroup$ 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 $\endgroup$ – Henry Oct 14 '16 at 17:10
  • $\begingroup$ 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. $\endgroup$ – Guido A. Oct 14 '16 at 17:11
  • $\begingroup$ @Henry The A would be $P$... the book uses different notation. I'll change it $\endgroup$ – SilenceOnTheWire Oct 14 '16 at 17:15
  • $\begingroup$ @SilenceOnTheWire OK, so $8$ ways of choosing the colour for the six faces coloured the same $\endgroup$ – Henry Oct 14 '16 at 17:17
  • $\begingroup$ @Henry yes. But why did you use $6!$ for the other colors? If you have 8 colors and use one, you still have 7*6*5*4*3*2 possible permutations, which is not $6!$... $\endgroup$ – SilenceOnTheWire Oct 14 '16 at 17:24

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