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Problem: Given 5 colors to choose from, how many ways can we color the four unit squares of a $2\times 2$ board, given that two colorings are considered the same if one is a rotation of the other?

I saw an existing thread on this question but the answers used Burnside's Lemma, but I am not familiar with Group Theory (and I don't think it is necessary for this problem).

I know how to do this problem if each square on the board must be a different color: There are $5*4*3*2$ possible colorings for the four unit squares and we just divide that amount by 4 to account for rotationally similar arrangements.

However, this problem implies that all squares can be the same color, so I am unsure how to do it.

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    $\begingroup$ "and we just divide that amount by $4$ to account for rotationally similar arrangements" Yes... that is what burnsides lemma tells you to do. You have been using burnsides lemma all along, though a perhaps simplified form for a special case. $\endgroup$ – JMoravitz Apr 28 '18 at 17:29
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    $\begingroup$ So, you've counted how many ways to do this if they were all different colors. Now... how about if there were three colors present (one was repeated twice). Count if the squares of the same color are adjacent (using the square on the left as a reference point), count if they are apart (no good reference point). Now, if only two colors appear, what if both colors appear twice? Together versus apart. If one color appears three times and the other once (use the color appearing once as a reference point). If all four squares are the same color. $\endgroup$ – JMoravitz Apr 28 '18 at 17:32
  • $\begingroup$ if you have, for example, the coloration $\begin{tabular}{ccc} A & B\\B & A \end{tabular}$, two rotations are the same. $\endgroup$ – Martín Vacas Vignolo Apr 28 '18 at 17:33
  • $\begingroup$ Thanks for the help! I solved it. $\endgroup$ – MathGuy Apr 28 '18 at 18:07

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