An office has 4 people in the Sales department, 3 in Accounts and 2 in Despatch. Three of the nine are chosen at random for a training programme. What is the probability that at least two of the three chosen come from the SAME department?

Count the opposite case, which is far simpler. Three people can be chosen in $C(9,3)=84$ ways. All three come from different departments in $4\times 3\times 2=24$ ways, one from each. So at least two share a department in 84 − 24 = 60 ways, a probability of 60/84 = 5/7. The value 2/7 = 24/84 is that complement left un-subtracted — it is the probability that all three departments are represented. The value 55/84 covers only 'exactly two from one department' and drops the 5 ways in which all three come from one department, namely $C(4,3)=4$ from Sales and $C(3,3)=1$ from Accounts; Despatch has only 2 people and so contributes none. The value 5/84 is that all-three-together case on its own. As a check, 24 + 55 + 5 = 84.