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AIME 1996: Interesting Counting Question

Two of the squares of a 7 by 7 checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane of the board. How many inequivalent color schemes are possible?

How would you do this AIME question with Polya-Burnsides Theorem (also called Burnsides Lemma) instead of direct counting or casework?

Note by Shaan Bhandarkar
4 years, 1 month ago

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Here's a link you might find helpful Taehyung Kim · 4 years, 1 month ago

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I separated it into two cases but then I used the theorem you mentioned (though I had never heard it called that before). I've basically just thought of it as intentional overcounting.

First case, the two squares are mirror images if viewed across the central square. (In this case, there are 24 pairs of these points). You will essentially be doublecounting these though if you count all 24 (because you could flip the board 90 degrees on half of them and get the other half). Thus, this case only has 12.

Second case, the two squares are not mirror images. Total number of ways to pick 2 out of 49 squares is 49C2 = (49)(48) / (2)(1). However, we cannot include the 24 pairs we just counted in case 1. Note, in this case, you actually count each true possibility 4 times (because the board has four sides you can view from). This means that we need to do [49C2 - 24] / 4 for this case.

Case 1 + case 2 = 12 + [49C2 - 24]/4 = 300 possibilities Owen Scott · 4 years, 1 month ago

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