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# mATH

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The sides of a regular decagon are to be colored with 5 different colors so that all 5 colors are used, and sides that are diametrically opposite will have the same color. One possible coloring is shown below.

[asy]
unitsize(2 cm);

draw(dir(36*0)--dir(36*1),linewidth(2*bp) + blue);
draw(dir(36*5)--dir(36*6),linewidth(2*bp) + blue);
draw(dir(36*1)--dir(36*2),linewidth(2*bp) + orange);
draw(dir(36*6)--dir(36*7),linewidth(2*bp) + orange);
draw(dir(36*2)--dir(36*3),linewidth(2*bp) + red);
draw(dir(36*7)--dir(36*8),linewidth(2*bp) + red);
draw(dir(36*3)--dir(36*4),linewidth(2*bp) + yellow);
draw(dir(36*8)--dir(36*9),linewidth(2*bp) + yellow);
draw(dir(36*4)--dir(36*5),linewidth(2*bp) + green);
draw(dir(36*9)--dir(36*0),linewidth(2*bp) + green);
[/asy]

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

Feb 18, 2020

#1
+1

Feb 18, 2020
edited by AnimalMaster  Feb 18, 2020
#2
+1

There are 5 ways to choose the sides for the first color, then 4 for the second, and so on.  That gives us 5*4*3*2*1 = 120 ways to color the decagon.

Feb 18, 2020
#3
+1

I don't think that's correct because you didn't cancel out rotations.

AnimalMaster  Feb 18, 2020
#4
+1

That's right, you must divide by 10, to get 120/10 = 12.

Guest Feb 18, 2020
#5
+1

I think it is just 4! = 24 ways

If you put 5 down on one side then the other 5 are automatically determined.

You want rotations to be the same so you need to one colour in one spot (any colour, any spot.)

Now there is 4 colours.

chose and place

now there is three, choose one and place it

etc

so that is

4*3*2*1 = 24 arrangements.

Feb 18, 2020
edited by Melody  Feb 18, 2020