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A friend has just shown me a question from his child's homework over which he's stumped (both child and father). Unfortunately I was unable to help as it stumped me, too. The question is...

There are...

4 starters

8 main meals

3 deserts

How many different 3 course meals can we make out of the above?

So how do we calculate the number of possible meals?

Thanks for your help.

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4 Answers 4

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Since you make independent choices, simply multiply the number of options for each course together, which gives $4\times 8 \times 3 = 96$.

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  • $\begingroup$ Thanks. I just used a slightly more convoluted method and came up with the same answer. Didn't think it would be as easy as just multiplying the numbers. Thanks for your help. $\endgroup$
    – Karl
    Sep 26, 2015 at 23:57
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As JWeissman states, the total number of three course meals, assuming each includes one starter, one main, and one dessert, would be $4\cdot 8 \cdot 3=96$. This uses the fundamental principle of counting. To see why the formula works, you could try making a tree diagram or listing possible outcomes. For a smaller example, consider just 2 appetizers and 3 mains. Let the appetizers be A & B and let the mains be C, D, and E. The possible combinations are: AC BC AD BD AE BE There are 2 choices for the first letter and 3 for the second, for a total of $2 \cdot 3 = 6$ choices.

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Think of a tree-like structure: take any option for the starter. There are 8 possible edges from it to main courses. For each main course there's an edge to 3 deserts. So for 1 starter you have 8*3 outcomes. And since you have 4 starters, then...

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How many different 3 course meal we make out? Let's say the chief simply called it x,

Let's take an example of two starters {salad, pasta} and three meals{boeuf bourguignon, boeuf à la plancha,roti de boeuf}.

We can then take: {salad,boeuf bourguignon},{salad, boeuf à la plancha}, {salad, roti de boeuf} {pasta, boeuf bourguignon},{pasta, boeuf à la plancha}, {pasta, roti de boeuf}. That it two say: two options

(hum, yummy yummy, I can choose either the first, or the second)

multiplied by three options

(wohoho, miam miam! I can choose either the first starter with one of the three options or the second starter with one of the three options! that is to say six different things!)

Generally speaking, there is always $x=\prod n_i$ with $n_i$ the number of things of one kind you may want to combine with the other $n_j$ things of other kind:

$$x=n_1*n_2*...*n_n$$

Hence the answer wich issimply multiply the number of options for each course together, which gives 4×8×3=... (you tell me!)

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