Wouldn't the 12 be for 1 single sector for zero? Seems to me that if we look at on a sector basis, the correct answer is 24. That is, we do know we must have 1 five, 1 three, and 2 zero's. Since the zero's are distinguishable by sectors should the zeros be distinguishable, e.g. a red zero and and a blue zero? If that is the case and we consider the arrangements different, i.e. (0,0,3,5) is different than (0,0,3,5), we have 24 possibilities. Please note, I'm asking - not telling.
WELL i CAN'T ALWAYS BE CORRECT: Just think of the 1 as a blue zero;
0, 0, 3, 5 == 0, 0, 5, 3 == 0, 1, 3, 5 == 0, 1, 5, 3 == 1, 0, 3, 5 == 1, 0, 5, 3 == 1, 1, 3, 5 == 1, 1, 5, 3
0, 3, 0, 5 == 0, 5, 0, 3 == 0, 3, 1, 5 == 0, 5, 1, 3 == 1, 3, 0, 5 == 1, 5, 0, 3 == 1, 3, 1, 5 == 1, 5, 1, 3
3, 0, 0, 5 == 5, 0, 0, 3 == 3, 0, 1, 5 == 5, 0, 1, 3 == 3, 1, 0, 5 == 5, 1, 0, 3 == 3, 1, 1, 5 == 5, 1, 1, 3
3, 0, 5, 0 == 5, 0, 3, 0 == 3, 0, 5, 1 == 5, 0, 3, 1 == 3, 1, 5, 0 == 5, 1, 3, 0 == 3, 1, 5, 1 == 5, 1, 3, 1
3, 5, 0, 0 == 5, 3, 0, 0 == 3, 5, 0, 1 == 5, 3, 0, 1 == 3, 5, 1, 0 == 5, 3, 1, 0 == 3, 5, 1, 1 == 5, 3, 1, 1
5, 3, 0, 0 == 3, 5, 0, 0 == 5, 3, 0, 1 == 3, 5, 0, 1 == 5, 3, 1, 0 == 3, 5, 1, 0 == 5, 3, 1, 1 == 3, 5, 1, 1
The bold is if there is no difference between the two zero's.