https://charlesreid1.com/w/index.php?action=history&feed=atom&title=FMM12FMM12 - Revision history2026-06-20T05:22:16ZRevision history for this page on the wikiMediaWiki 1.39.12https://charlesreid1.com/w/index.php?title=FMM12&diff=27047&oldid=prevAdmin: Created page with "{{FMM |title=Checkerboard Color Schemes |problem= Two of the squares of a 7 x 7 checkerboard are painted yellow, and the rest are painted green. Two color schemes are equival..."2019-11-16T22:39:03Z<p>Created page with "{{FMM |title=Checkerboard Color Schemes |problem= Two of the squares of a 7 x 7 checkerboard are painted yellow, and the rest are painted green. Two color schemes are equival..."</p>
<p><b>New page</b></p><div>{{FMM<br />
|title=Checkerboard Color Schemes<br />
|problem=<br />
<br />
Two of the squares of a 7 x 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?<br />
<br />
Hint: There are <math>\binom{49}{2} = 1176</math> ways to select the positions of the yellow squares. However, because we can apply quarter-turns, there are less than 1176 inequivalent color schemes.<br />
<br />
|solution=<br />
<br />
Color schemes fall into two classes:<br />
<br />
1. color schemes in which the two yellow squares are not diametrically opposed<br />
<br />
2. color schemes in which the two yellow squares are diametrically opposed<br />
<br />
Case (1) appears in four equivalent forms, so we will divide the total number of color schemes in Case (1) by 4.<br />
<br />
Case (2) appears in two equivalent forms,so we will divide the total number of color schemes in Case (2) by 2. We also know there are <math>\dfrac{49 - 1}{2} = 24</math> such pairs of yellow squares, contributing (24/2) total inequivalent color schemes.<br />
<br />
The number of cases in the Case (1) class is the total number of arrangements, minus the 24 pairs of yellow squares in the Case (2) class - and we divide it by 4, so Case (1) contributes (1176 - 24)/4 total inequivalent color schemes.<br />
<br />
The total is therefore:<br />
<br />
<math><br />
\dfrac{1176 - 24}{4} + \dfrac{24}{2} = 300<br />
</math><br />
<br />
<br />
}}</div>Admin