 Solution

There are $6^3=216$ unrestricted placements, since one colour has $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ occupancy vectors. For any specified empty box, the two remaining boxes admit $3^3=27$ placements. For any specified pair of empty boxes, the one remaining box admits one placement. All three boxes cannot be empty.

The inclusion-exclusion principle therefore givesThe empty-box events are not disjoint events, so subtracting their counts without restoring their pairwise overlaps would be incorrect.