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If all the 20 faces of a regular icosahedron are painted with a set of 20 distinct colours then the total number of such icosahera possible. The cube analogue of this is more well known and the answer to it is 30.

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  • $\begingroup$ Does "distinct colors" mean that we are not allowed to use the same color on two different faces? $\endgroup$ – lisyarus Jan 18 at 15:13
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Just like the cube version - take the symmetric group on the colors (order $20!$) and then divide by the rotations of the regular icosahedron ($A_5$, order $60$). Overall, that's $\frac{20!}{60}=\frac{19!}{3}$.

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    $\begingroup$ Could you elaborate the terms in your answer. The solution I know to the cube was completely different. $\endgroup$ – Karthik Jan 18 at 14:06
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    $\begingroup$ Answering to your elaboration quest: there are 20! ways to arrange 20 colors: this is the set of possible permutations. But then the icosahedron itself has a symmetry group (way of orientations) of order 60. That is, 60 colorings each are to be considered identical because there is some way to rotate the icosahedron into an already counted coloring. Therefore you have 20!/60 = 19!/3 = 19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*2*1 = 40 548 366 802 944 000 ways to color an icosahedron. $\endgroup$ – Dr. Richard Klitzing Jan 18 at 14:47
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    $\begingroup$ It's a good interpretation of the Question, but it would improve the Answer to state the critical assumptions, that each of the 20 colors is used exactly once (from which it follows that the same coloring could occur from rotation only when the rotation is trivial). $\endgroup$ – hardmath Jan 18 at 16:43

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