Five different letters are to be dropped into three different post boxes, each letter going into exactly one box. In how many ways can this be done if no box is left empty?

With no condition, each of the 5 letters can go into any of the 3 boxes, giving $3^5=243$ ways. Now strike out the ways that leave a box empty. Ways that avoid one named box use only the other two: $2^5=32$, and there are 3 boxes that could be the empty one, so $3\times 32=96$ to remove. But the ways that use only ONE box have been struck twice over, and there are 3 of those, so they must be added back: 243 − 96 + 3 = 150. Trap 243 forgets the condition. Trap $125=5^3$ reverses the roles — it is the count for 3 letters going into 5 boxes. Trap $60=5\times 4\times 3$ seeds one letter into each box and then forgets that the remaining two letters still have to go somewhere.