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!


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: ${5 \choose 0}+{5 \choose 1}+{5 \choose 2}$

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

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