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A cerntain men's club has sixty members; thirty are business men and thirty are professors. In how many ways can a committee of eight be selected if at least three must be business men and at least three professors?

I have the solution to this. The way they did it is to add up the cases in which you have 3 professors and 4 professors and 5 professors to get $$2{30 \choose 3}{30 \choose 5} + {30 \choose 4}^2$$

I understand that answer, but I thought about it a different way. My thought process was that there should be 3 that definitely have to be professors and there are ${30 \choose 3}$ ways to select them. There are also 3 that definitely have to be business men and there are ${30 \choose 3}$ ways to select these guys as well. There are two positions left and they must come from the remaining 54 members so I get a total of $${30 \choose 3}^2{54 \choose 2}$$ ways of arranging the 8. Why is this incorrect?

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    Because you've multiple-counted some of the combinations. Let's just focus on the professors, which we'll call A through Z. If the eventual selection contains professors A, B, D, E, G, then you'll count that combination $10$ times: once when your first three are A, B, D; once when they're A, B, E; once when they're A, B, G; once when they're A, D, E; and so on. Same with the businessmen. – 2017-02-10

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Short answer: You are over counting.

Long answer:

$\binom {30}3^2\binom{54}{2}$ counts ways to select 2 from 30 business men and 2 from 30 professors into subgroup A, and any 2 from the remaining 52 personages in subgroup B. Which is okay if the committee is so divided, but it is not.

Selecting professors {Adam, Ben, Carl}, then {Donna}, or {Adam, Ben, Donna} then {Carl}, and so on, result in the same committee and should not be counted four times over. Similarly for selecting business men.