Out of the n colors you could have chosen 1, 2, 3 or 4 colors to paint the tetrahedron.
There are n ways of choosing 1 color.
There are C(n,2) ways of choosing 2 colors, but then 3 ways of choosing whether color A is on 1, 2 or 3 faces.
There are C(n,3) ways of choosing 3 colors, but then 3 ways of choosing which one is to be duplicated.
There are C(n,4) ways of choosing 4 colors. This is the only case where left- and right-handed versions exist, and so needs to be multiplied by 2.
In total for the tetrahedron: n + 3*C(n,2) + 3*C(n,3) + 2*C(n,4)
The solution below for the cube is INCORRECT. See later discussion.
For the cube:
You could choose to paint any given cube with 1, 2, 3, 4, 5 or 6 colors.
There are n ways of choosing 1 color.
There are C(n,2) ways of choosing 2 colors. Color A can then appear on 1, 2, 3, 4 or 5 faces. If on 1 or 5 faces, there is only 1 way of arranging them. If on 2 or 4 faces, the minority color (A or B) can be on adjacent or opposite faces, for 2 ways each. If color A is on 3 faces, they can be gathered around a vertex or include two opposite faces and one connecting face. This then accounts for 1+2+2+2+1=8 ways for each of the C(n,2) choices of color.
There are C(n,3) ways of choosing 3 colors. Within this, the colors can be distributed as 1, 2 and 3, or as 2,2,2 or as 1,1,4. The ways of assigning colors A, B and C to these numbers is 3!=6, 1 and 3 respectively. Then also, if the distribution is 1,2,3, the tripled-up color can be on three faces in a row or on three faces adjoining the same vertex; if the latter, there's no further breakdown, but if the former, the singlet color can be either opposite the middle face of the triplet or opposite one of the doublet's two faces. If the distribution is 2,2,2, there is just a left- and a right-handed version. If the distribution is 1,1,4, the two singlets can be either opposite or adjacent.
So for 3 colors the ways are 3*3! + 2 + 2*3 = 26 for each of the C(n,3) ways of choosing the 3 colors.
There are C(n,4) ways of choosing 4 colors. The four chosen colors can be distributed 1,1,1,3 or 1,1,2,2. The ways of assigning the colors A, B, C, D to the possibilities are 4 and C(4,2)=6 respectively. In each of these cases: With 3 of one color, the 3 can be arranged in a row or around one vertex; if the former, then any of 3 colors can be opposite the middle square of the triplet; if the latter, there are 2 mirror-image ways of placing the other 3 colors. This amounts to 5 ways of arranging each of the 4 color assignments for 1,1,1,3.
If, however, the 4 colors are arranged 1,1,2,2, there are three possibilities: If the two doubled-up colors are both arranged opposite their matching squares, there's only one version. If one doublet is adjacent but the members of the other pair are opposite, the is also only one version. But if both doublets are arranged so that matching colors are adjacent, there's a left-handed version and a right-handed version, just considering the doublets, and then there are 2 ways for each of these in regard to the placement of the remaining two colors.
So for 4 colors there are (2+3)*4 + (4+1+1)*C(4,2) = 60 ways for each of the C(n,4) ways of choosing the 4 colors.
There are C(n,5) ways of choosing 5 colors. For each of these there are 5 ways of choosing which is to be appear twice. If the matching faces (color A) are opposite, there are 3 ways of arranging the remaining four colors (any of C, D or E can be opposite B). If the matching faces are adjacent, there are C(4,2) ways of choosing which 2 of the 4 remaining colors are adjacent to both members of the maching pair, and then 2 ways of choosing the placement of the other two colors. So there are 5*(3+C(4,2)*2)=75 ways for each of the C(n,5) ways of choosing 5 colors.
There are C(n,6) ways of choosing 6 colors. There are then 5 choices for the color to be opposite color A. For each of these there are 6 ways of choosing the order for the remaining 4 colors. So there are 5*6=30 ways for each of the C(n,6) ways of choosing the 6 colors.
That all adds up to n + 8*C(n,2) + 26*C(n,3) + 60*C(n,4) + 75*C(n,5) + 30*C(n,6) ways of painting the cube all together.
The above solution for the cube is incorrect. See subsequent discussion.
Edited on September 25, 2004, 5:41 pm
Posted by Charlie
on 2004-09-25 14:38:17