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The edges of a square are to be colored either red, blue, yellow, pink, or black. Each side of the square may only have one color, but a color may color many sides. How many different ways are there to color the square if two ways that can obtained from each other by rotation are identical?

So, it seems to simply be a variation of a circular permutation problem. The only real complication is the fact that there are 5 colors to choose from in a 4 position arrangement.

My solution to this is $$3!(5C4)+\frac{3!(5C3)}{2!}+\frac{3!(5C2)}{2!2!}+\frac{3!(5C2)}{3!}+5=90$$

The denominators account for the repetition of colors to fill all 4 sides when only 3 or 2 unique colors are selected.

While this was my initial solution, the answer given to us turned out to be 165. How is this?

SolvingTrainee
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Your first and last terms are correct, but rest are not. There are two ways of coloring when $3$ colours are to be used (as shown in this picture). Either the same coloured edges are opposite or are adjacent.enter image description here Accordingly, number of ways of colouring are $5C1\times4C2$ and $5C1\times4C2\times2$, i.e. $30$ and $60$ ways.

There are three ways of colouring when $2$ colours are to be used (as shown in the following picture).enter image description here Accordingly, number of ways of colouring are $5C2\times2$, $5C2$ and $5C2$ respectively. So, total number of ways of colouring are $$(30)+(30+60)+(20+10+10)+(5)=\boxed{165}$$

Martund
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