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When a customer buys a familysized meal at certain restaurant, they get to choose 3 side dishes from 9 options. Suppose a customer is going to choose 3 different side dishes.
How many groups of 3 different side dishes are possible?

When a customer buys a familysized meal at certain restaurant, they get to choose $3$ side dishes from $9$ options. Suppose a customer is going to choose $3$ different side dishes. $\newline$ How many groups of $3$ different side dishes are possible?

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Counting principle

Full solution

Q. When a customer buys a familysized meal at certain restaurant, they get to choose

3

side dishes from

9

options. Suppose a customer is going to choose

3

different side dishes.

\newline

How many groups of

3

different side dishes are possible?

Identify problem type: Identify the problem type. $\newline$ We need to find the number of combinations of $3$ side dishes that can be chosen from $9$ options. This is a combination problem because the order in which the side dishes are chosen does not matter.
Use combination formula: Use the combination formula. $\newline$ The number of ways to choose $3$ items from $9$ is given by the combination formula: $C(n, k) = \frac{n!}{k!(n-k)!}$ , where $n$ is the total number of items to choose from, $k$ is the number of items to choose, ＂ $!$ ＂ denotes factorial, and $C(n, k)$ is the number of combinations.
Calculate number of combinations: Calculate the number of combinations. $\newline$ Using the combination formula with $n=9$ and $k=3$ , we get: $\newline$ $C(9, 3) = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9\times8\times7}{3\times2\times1} = \frac{504}{6} = 84$ .

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