We worked on two different cube coloring problems before. One is to paint a unit cube with $1$ face in red, $1$ face in green, $1$ face in yellow, and $3$ faces in blue color. The other is to paint a unit cube with $2$ faces each in red, green and blue.<\/p>\n\n\n\n
By using P\u00f3lya enumeration theorem<\/a> or Bunside’s lemma<\/a>, the number of ways with $m$ colors to paint a cube can be calculated with the following formula: $$c(m)=\\dfrac{1}{24}(m^6+3m^4+12m^3+8m^2)$$ The above formula includes coloring with $1$ color, $2$ colors, …, $m$ colors. For example, $$c(2)=10\\ \\ \\ \\ \\ \\ \\ \\ \\ c(3)=57$$ The ways to pain the cube with 6 different colors and each face having a different color is $\\dfrac{6!}{24}=30$ because the Rotational Symmetries of the Cube<\/a> is 24.<\/p>\n","protected":false},"excerpt":{"rendered":" We worked on two different cube coloring problems before. One is to paint a unit cube with $1$ face in red, $1$ face in green, $1$ face in yellow, and $3$ faces in blue color. The other is to paint … Continue reading