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A committee must be formed with 2 teachers and 3 students. If there are 9 teachers to choose from, and 9 students, how many different ways could the committee be made?
Answer:

A committee must be formed with $2$ teachers and $3$ students. If there are $9$ teachers to choose from, and $9$ students, how many different ways could the committee be made? $\newline$ Answer:

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Counting principle

Full solution

Q. A committee must be formed with

2

teachers and

3

students. If there are

9

teachers to choose from, and

9

students, how many different ways could the committee be made?

\newline

Answer:

Choose $2$ teachers: Determine the number of ways to choose $2$ teachers out of $9$ . $\newline$ We use the combination formula, which is $C(n, k) = \frac{n!}{k!(n-k)!}$ , where $n$ is the total number of items to choose from, $k$ is the number of items to choose, and ＂!＂ denotes factorial. $\newline$ For choosing $2$ teachers out of $9$ , we have: $\newline$ $C(9, 2) = \frac{9!}{2!(9-2)!} = \frac{9!}{2!7!} = \frac{9 \times 8}{2 \times 1} = 36$
Choose $3$ students: Determine the number of ways to choose $3$ students out of $9$ . $\newline$ Similarly, we use the combination formula. $\newline$ For choosing $3$ students out of $9$ , we have: $\newline$ $C(9, 3) = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$
Calculate total ways: Calculate the total number of ways to form the committee. $\newline$ Since the selection of teachers and students are independent events, we multiply the number of ways to choose the teachers by the number of ways to choose the students. $\newline$ Total number of ways $=$ Number of ways to choose teachers $\times$ Number of ways to choose students $\newline$ Total number of ways $= 36 \times 84$ $\newline$ Total number of ways $= 3024$

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