A three-member panel is picked at random from a group of 4 economists, 3 statisticians and 2 lawyers. What is the probability that the panel ends up holding one person of each kind?

The group has 4 + 3 + 2 = 9 people, so total panels = $C(9,3)=84$. A panel with one of each kind is built by taking any economist, any statistician and any lawyer: $4\times 3\times 2=24$. Probability = 24/84 = 2/7. Trap 3/28 = 9/84 ADDS the group sizes, 4 + 3 + 2 = 9, where they must be multiplied — the three choices are made alongside one another, not instead of one another. Trap 1/21 = 24/504 pairs the unordered numerator 24 with the ORDERED total $9\times 8\times 7$; numerator and denominator have to be counted the same way. Trap 1/7 = 12/84 multiplies only the first two groups and forgets the lawyers.