
EXAMPLE

For n=3 the 20 different arrangements of 1,1,1,1,1,1 result in 7 energy levels (sum of signed inverse distances):
{0,0,0,1,1,1},{1,1,1,0,0,0}: 13/10
{0,0,1,0,1,1},{1,1,0,1,0,0}: 41/30
{0,0,1,1,0,1},{0,1,0,0,1,1},{1,0,1,1,0,0},{1,1,0,0,1,0}: 56/30
{0,0,1,1,1,0},{0,1,1,1,0,0},{1,0,0,0,1,1},{1,1,0,0,0,1}: 8/10
{0,1,0,1,0,1},{1,0,1,0,1,0}: 37/10
{0,1,0,1,1,0},{0,1,1,0,1,0},{1,0,0,1,0,1},{1,0,1,0,0,1}: 89/30
{0,1,1,0,0,1},{1,0,0,1,1,0}: 71/30
so the multiplicities are 4*2 + 3*4 = 20 = binomial(6,3).
