It is known that there are eight nonequivalent colorings of the corners of a regular pentagon with the colors red and blue. Explicitly determine eight nonequivalent colorings.

Using Burnside's Lemma I found that there exist exactly eight nonequivalent colorings for a pentagon using 2 colors.

I know how to find different permutations of a certain coloring to get equivalent colorings but how do I find the nonequivalent colorings?

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    $\begingroup$ Just get out your crayons and try it. $\endgroup$ – Hagen von Eitzen Dec 8 '13 at 21:31

A good way of finding all possibilities is to distinguish cases which are clearly different. In this example we can take the number of red sides as such a distinguishable property (as two colorings are always nonequivalent if they haven't the same number of red corners):

0 corners red: 1 possibility
1 corners red: 1 possibility (they are equivalent by rotation)
2 corners red: 2 possibilities (1 when the red dots are neighbours and 1 when they are not)
3 corners red: 2 possibilities (1 when the blue dots are neighbours and 1 when they are not)
4 corners red: 1 possibility
5 corners red: 1 possibility Total: 8 possibilities

Note that it wasn't necessairely to do the work for 3, 4 and 5 red corners as we could have noticed that there should be 2, 1 and 1 possibility by symmetry (as m red corners and n blue corners should give the same number of possibilities as n red and m blue corners).


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