PUMaC 2013 · 组合（A 组） · 第 2 题

PUMaC 2013 — Combinatorics (Division A) — Problem 2

[ 3 ] How many ways are there to color the edges of a hexagon orange and black if we assume that two hexagons are indistinguishable if one can be rotated into the other? Note that we are saying the colorings OOBBOB and BOBBOO are distinct; we ignore flips.

解析

[ 3 ] How many ways are there to color the edges of a hexagon orange and black if we assume that two hexagons are indistinguishable if one can be rotated into the other? Note that we are saying the colorings OOBBOB and BOBBOO are distinct; we ignore flips. Solution The rotation group for the hexagon consists of the identity, two rotate-by- π/ 3 oper- ations, two rotate-by-2 π/ 3, and one rotate by π operation. There are 64 colorings fixed by the identity, 2 by the two rotate-by- π/ 3 operations, 4 by the two rotate-by-2 π/ 3 operations, and 8 by the rotate-by- π operation. Then the number of distinct colorings is given by Burnside’s 1 84 lemma to be (1(64) + 2(2) + 2(4) + 1(8)) = = 14. So there are 14 colorings. 6 6