Problem:
Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?
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Solution:
Let , and be the number of red, white, and blue faces, respectively. Then is one of 15 possible ordered triples, namely one of the three permutations of , or , or one of the six permutations of . The number of distinguishable colorings for each of these ordered triples is the same as for any of its permutations. If , then exactly one coloring is possible. If , the tetrahedron can be placed with the white face down. If , the tetrahedron can be placed with one white face down and the other facing forward. If , the tetrahedron can be placed with the white face down and the blue face forward. Therefore there is only one coloring for each ordered triple, and the total number of distinguishable colorings is .
The problems on this page are the property of the MAA's American Mathematics Competitions