You have an unlimited supply of wooden regular tetrahedra and N cans of paint, each of a different color.
You want to paint as many tetrahedra as possible given that you're limited to these N colors, with one color per face, but not requiring different colors on different faces, so that no two tetrahedra are identical. Two tetrahedra can count as non-identical even if they are mirror images, reversed.
It turns out that, given this value of N, you can make exactly as many tetrahedra with three colors as you can with four colors.
What's the value of N, the number of different colors available?
How many different tetrahedra do you have all together?
With N colors, a 3 color painting can be done in N*(N-1)*(N-2) different ways. One color must necessarily be on two sides, and there are N choices. Once painted, position the tetrahedron so that the colored sides are on the bottom and the back side. There are N-1 ways to paint the left face, after which there are N-2 ways to paint the right.
Also, a 4-color painting can be done in N*(N-1)*(N-2)*(N-3)/6 ways. Number the sides 1 - 4, and then there are N ways to paint side one, and then N-1 ways to paint side 2, etc. But this overcounts the number of unique paintings by a multiple of 6, because any 4 colors can be painted in 12 different sequences (4*3*2*1) but there are only two distinct colorings (the reflections)
If the two are equal, then (N-3)/6 = 1, or N = 9
And the number of different tetraheda colorings = 9*8*7 = 504
Edited on August 22, 2011, 6:15 pm
I think .. (spoiler)
