Maths Olympiad Prep

Track / Stage 5 / 182 of 400 #782 of 1964

Problem 782

AIME late
Combinatorics Difficulty 5.5 Find the answer

Example 4.4.3 Color the six faces of a cube with three colors: red, blue, and green. How many different coloring schemes are there? How many coloring schemes are there if each color appears exactly twice?

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

Solution: At this point, D={1,2,3,4,5,6},R={r,b,g}D=\{1,2,3,4,5,6\}, R=\{r, b, g\}.
The following discusses the permutation group GG of DD, which has 24 elements:
(1) The identity permutation type is 161^{6}, with 1 element;
(2) The permutation type for a 90,27090^{\circ}, 270^{\circ} rotation around the axis through the centers of opposite faces is 12411^{2} 4^{1}, with 6 elements; the permutation type for a 180180^{\circ} rotation is 12221^{2} 2^{2}, with 3 elements;
(3) The permutation type for a 120,240120^{\circ}, 240^{\circ} rotation around the axis through opposite vertices is 323^{2}, with 8 elements;
(4) The permutation type for a 180180^{\circ} rotation around the axis through the midpoints of opposite edges is 232^{3}, with 6 elements; thus,
PG(x1,x2,x3,x4,x5,x6)=124(x16+6x12x4+3x12x22+8x32+6x23). P_{G}\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}\right)=\frac{1}{24}\left(x_{1}^{6}+6 x_{1}^{2} x_{4}+3 x_{1}^{2} x_{2}^{2}+8 x_{3}^{2}+6 x_{2}^{3}\right).
The number of equivalence classes is
N(G,C)=PG(3,3,3,3,3,3)=124(36+6×33+3×34+8×32+6×33)=57,FG(i=13wi,i=13wi2,,i=13wi6)=124[(r+b+g)6+6(r+b+g)2(r4+b4+g4)+3(r+b+g)2(r2+b2+g2)2+8(r3+b3+g3)2+6(r2+b2+g2)3]. \begin{array}{l} N(G, C)=P_{G}(3,3,3,3,3,3) \\ =\frac{1}{24}\left(3^{6}+6 \times 3^{3}+3 \times 3^{4}+8 \times 3^{2}+6 \times 3^{3}\right)=57, \\ F_{G}\left(\sum_{i=1}^{3} w_{i}, \sum_{i=1}^{3} w_{i}^{2}, \cdots, \sum_{i=1}^{3} w_{i}^{6}\right) \\ =\frac{1}{24}\left[(r+b+g)^{6}+6(r+b+g)^{2}\left(r^{4}+b^{4}+g^{4}\right)+3(r+b+g)^{2}\right. \\ \left.\quad\left(r^{2}+b^{2}+g^{2}\right)^{2}+8\left(r^{3}+b^{3}+g^{3}\right)^{2}+6\left(r^{2}+b^{2}+g^{2}\right)^{3}\right] . \end{array}

The coefficient of r2b2g2r^{2} b^{2} g^{2} in the above expansion is
124(6!2!2!2!+36+63!1!1!1!)=6. \frac{1}{24}\left(\frac{6!}{2!2!2!}+3 \cdot 6+6 \cdot \frac{3!}{1!1!1!}\right)=6 .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.