On 21/06/2016 01:44, Keith F. Lynch wrote:
Can anyone think of a practical way to count the arrangements? The obvious way would be to try all 2^60 arrangements and eliminate the ones that are rotations of arrangements that have already been found. But 2^60 is more than 10^18, so that's just barely computationally feasible. Is there a better way? Has someone counted them already?
You're looking for the Polya enumeration theorem. I always find this painfully fiddly to apply and am likely to get the following wrong, but it goes like this. The number of configurations is the average, over all elements of the symmetry group, of m^c where m is the number of "colours" (2, in this case) and c is the number of cycles in the permutation produced by the group action. In this case our group is A_5 and it goes like this: - the identity has 60 (trivial) cycles - there are 15 elements with 30 cycles of length 2 - there are 20 elements with 20 cycles of length 3 - there are 24 elements with 12 cycles of length 5 giving 1/60 (2^60 + 15.2^30 + 20.2^20 + 24.2^12) different configurations. If I haven't made a mistake, which I probably have. -- g