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I've used this site a lot for help understanding problems in my other math classes. Although I never got around to actually asking a question since most were already asked. Today it's a simple one involving combinatorics. Admittedly, I'm not very good with combinatoric problems, but they've really only been mentioned as an aside in my other classes. Well I'm finally taking a probability course and the first section is on combinatorics. The question I'm having trouble with follows:

A salad bar has 3 choices of greens, 8 veggies, 5 fruits, 3 dairy items, and 4 dressings. In how many ways can I serve myself a salad with at least one of the greens, either 3 or 4 veggies, 2 dairy items, no more than 2 fruits, and exactly one dressing?

I understand most of the question, but what really gets me are the "either" and "no more than." I know that if it was to select 4 veggies then it would be ${8 \choose 4}$, but since it's 3 or 4, I'm having some trouble figuring out what to do. My reasoning right now is to do ${4 \choose 3}{8 \choose 4}$ because there are that many ways to choose 3 out of four from choosing the 8 out of four. Is that correct?

Similarly, I've been stuck on the no more than 2 fruits part and have been interpreting that as 0, 1, or 2 fruits. So going by the same rationale would be ${1 \choose 0}{2 \choose 1}{5 \choose 2}$. Is my reasoning correct?

Once again, I understand the rest of the parts, just these two are giving me trouble. Thanks!

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Either three or four means you may have three and you may have four. In this case the choices are mutually exclusive. The number of ways to do that is ${8 \choose 4}+{8 \choose 3}$ For the fruits, you are correct that you might have $0,1,2$, but there is no choosing within that, so the same logic applies: ${5 \choose 0}+{5 \choose 1}+{5 \choose 2}$

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  • $\begingroup$ Thank you, that makes sense. $\endgroup$ – lmn123 Sep 10 '13 at 19:33

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