(In reply to solution: much opportunity for error
Corrected a few minor errors after seeing Charlie's self-response:
"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....If the distribution is 2,2,2, there is just a left- and a right-handed version..."
I disagree with this assertion. In the 2,2,2 case with colors A,B,C, we have the following possibilities:
- Each of the pairs (A,A'), (B,B'), and (C,C') are on opposite faces.
- One of the pairs is on opposite faces. Since there are three colors, this represents 3 possibilities.
- Each of the pairs (A,A'), (B,B'), and (C,C') are on adjacent faces. In this case, we have the right-handed and left-handed versions for 2 possibilities. (To see this more easily, position the cube in the unique way so that (A,A') are represented by (front,top) and so that the right face is B. There are two places for B'.)
This makes for 3*3! + 6 + 2*3 = 30 (not 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...
"If, however, the 4 colors are arranged 1,1,2,2, there are three possibilities...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."
I also disagree with this analysis. When both doublets are adjacent, we either have that the two leftover faces are adjacent or opposite. If opposite, there is just one possibility. If adjacent, orient the cube in the unique position such that (A,A') is represented by (front,top) and the right face is color B. There are 2 places to place the remaining B' and 2 ways to place the remaining C and D. This makes for 5 possibilities.
This makes for (2+3)*4 + (5+2+1)*C(4,2) = 68 (not 60) ways for each of the C(n,4) ways of choosing the 4 colors.
(Added note: the factor of 1 in front of C(4,2) was changed to a 2, since the case of one doublet being adjacent and one doublet being opposite actually amounts to two cases)
As a result, my final formula is:
n + 8*C(n,2) + 30*C(n,3) + 68*C(n,4) + 75*C(n,5) + 30*C(n,6) ways of painting the cube all together.
Edited on September 25, 2004, 5:22 pm