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Of a company's personnel, seven work in design, 14 in manufacturing, four in testing, five in sales, two in accounting, and three in marketing. A committee of six people is to be formed to meet with upper management. In how many ways can the committee be formed if there must be at least two members from the manufacturing department?

Am I double counting if I have $\left.14\choose 2\right.\left.33\choose 4\right.$? The logic I'm using is that once your 2 people are chosen for the two required manufacturing seats, we don't care what other departments fill the other 4 seats. But I've been told in some other problems (of which this seems similar) that I could be double counting if I do it like this.

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You are. It is allowable to have A,B,C from manufacturing, and D,E,F from elsewhere. You will count that three times, once with each pair being the first two from manufacturing.

The easiest way I see for this problem is to take all committees and subtract those with nobody from manufacturing and those with exactly one, so ${35 \choose 6}-{21 \choose 6}-{14 \choose 1}{21 \choose 5}$

Another way is to add up the number of committees with exactly each allowed number from manufacturing, so two would be ${14 \choose 2}{21 \choose 4}$

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Yes you are. E.g. if everyone came from manufacturing, you're counting each instance of this situation $\binom{6}{2}$ times (counting once for each pair of manufacturers which end up counted by the first "choose").

I'd use a variable to count the number of representatives from manufacturing, e.g. $i$, and sum over it: So the number is $$\sum_{i=2}^6 \text{number of possible committees with } i \text{ representatives from manufacturing}.$$

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