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I don't know how to count

 

In a regular decagon, the sides are to be colored with five different colors, so that sides are diametrically opposite have the same color.  (Not all five colors need to be used.)  One possible coloring is shown below.

 

In how many different ways can the sides of the decagon be colored? (Two colorings are considered identical if one can be rotated to form the other.)

 Feb 11, 2023
 #2
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Try to use Burnside's lemma 

 Feb 11, 2023
 #3
avatar+118616 
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There is really only 5 sides to be set because once you have one you automatically have its opposite.

Also roations are the same

so I think it is just 

4! if all the colours are used  that is  24 ways

 

If only 4 colours are used then I think it is  5C4 * 4C1 * 4!/2     or    5C4 * 4C1 * 

1  =  5*4*12 = 240 ways

 

If only 3 colours are used and 3 are the same then maybe it is  5C3 * 3C1 * 4!/3! = 10 * 3 * 4 = 120ways

 

If 3 colours are used and 2 have 2 each then   i'm going for     5C3 * 3C2 *  4!/(2!2!) = 10*3*6 = 180 ways

 

If 2 colours are used then you have to look at each senario that is  1:4,  2:3  splits

1:4,   split      5C2*2 * 2C1 * 1 = 10 * 2 = 20 ways

2:3    split       5C2* 2   * 4C1 = 10* 2* 4 = 80 ways

 

If only 1 colour is used that there is only 1 possibility.   

 

24+240+120+80+20+80+1 = 565

 

 

It is highly likely that one or more of these are wrong.

 Feb 13, 2023

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