Nine chairs stand in a straight row. Four guests are to be seated on them so that no two guests occupy neighbouring chairs, the other five chairs staying empty. In how many ways can the four guests be seated?

Put the five empty chairs down first. They create 6 gaps — one before, four between and one after — and no two guests are neighbours exactly when each guest takes a different gap. Choose 4 of the 6 gaps in $C(6,4)=15$ ways, then seat the four different guests in those gaps in 4! = 24 ways: $15\times 24=360$. Trap $3024=9\times 8\times 7\times 6$ is the unrestricted count and ignores the no-neighbours rule. Trap 15 stops at choosing WHICH chairs are used and treats the four guests as interchangeable, which they are not. Trap 2880 = 4! × 5! treats the five empty chairs as five different objects to be arranged, which both over-counts them and drops the condition.