Set A can have two members in either of 4 ways:
{0,7}{1,6}{2,5}{3,4}
In each of these 4 possibilities, B and C can be divided with B getting 1, 2 or 3 out of the 8 remaining.
So far we get 4 * (C(8,1)+C(8,2)+C(8,3))
Set A can have four members in any of 13 ways:
{0,1,4,9}{0,1,5,8}{0,1,6,7}{0,2,3,9}{0,2,4,8}{0,2,5,7}{0,3,4,7}{0,3,5,6}
{1,2,3,8}{1,2,4,7}{1,2,5,6}{1,3,4,6}{2,3,4,5}
B can have 1 or 2 of the remaining 6.
So we add in 13 * (C(6,1)+C(6,2))
Set A can have six members in 9 ways:
{0,1,2,3,6,9}{0,1,2,3,7,8}{0,1,2,4,5,9}{0,1,2,4,6,8}{0,1,2,5,6,7}{0,1,3,4,5,8}{0,1,3,4,6,7}{0,2,3,4,5,7}{1,2,3,4,5,6}
Set B can have only 1 of the remaining four:
So add 9 * C(4,1)
When set A has 8 elements, B and C can't be divided unevenly without an empty set.
4 * (C(8,1)+C(8,2)+C(8,3)) + 13 * (C(6,1)+C(6,2)) + 9 * C(4,1) = 677
final (?) revision
