Three fair dice are rolled together. What is the probability that no two of the three dice show the same number?

Total outcomes = $6^3=216$. For all-different faces the first die has 6 choices, the second 5 and the third 4: $6\times 5\times 4=120$. Probability = 120/216 = 5/9. Trap $5/54=C(6,3)/216=20/216$ counts the sets of three faces but forgets that the three dice can carry those faces in 3! different ways. Trap 4/9 is the complement, 'at least two dice match'. Trap 1/36 = 6/216 is 'all three dice show the same number'.