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Anna is trying to decide what to have for lunch in the cafeteria. She can choose 1 entree and 2 side dishes. There are 4 available entrees, and 8 available side dishes. How many different combinations are ossible for Anna's lunch?

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    $\begingroup$ can she skip lunch entirely, or choose less than 3 dishes, or choose a double helping of a dish? $\endgroup$ – Will May 15 '14 at 17:54
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You have $4$ entrees to pick from, and you choose 1. $$4\choose1$$ You have 8 side dishes to pick from, and you choose 2. $$8\choose2$$ Put those together to get $$\binom{4}{1}\binom{8}{2}=4\cdot28=112$$

Keep in mind that this assumes the two side dishes must be different. If they can be the same, then there are $8\choose1$ ways to pick a doubled up sidedish. You can add this to the original $8\choose2$ for all your side dish combinations, then multiply by your entree possibilities for a new answer.

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    $\begingroup$ Adding to this, they can be put together because these two event are independent of each other. $\endgroup$ – fahrbach May 15 '14 at 17:58
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If she must choose exactly $1$ entree and exactly $2$ unique side-dishes, then:

  • Choose $1$ out of $4$ items: $\binom{4}{1}=\frac{4!}{1!\times3!}=4$
  • Choose $2$ out of $8$ items: $\binom{8}{2}=\frac{8!}{2!\times6!}=28$
  • So she can make $4\times28=112$ different combinations
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