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A committee must be formed with 3 teachers and 5 students. If there are 11 teachers to choose from, and 12 students, how many different ways could the committee be made?
Answer:

A committee must be formed with $3$ teachers and $5$ students. If there are $11$ teachers to choose from, and $12$ 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

3

teachers and

5

students. If there are

11

teachers to choose from, and

12

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

\newline

Answer:

Calculate Teachers Combinations: To determine the number of different ways to form the committee, we need to calculate the combinations of teachers and students separately and then multiply them together. For the teachers, we will 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.
Calculate Students Combinations: First, we calculate the number of ways to choose $3$ teachers from $11$ . Using the combination formula, we get $C(11, 3) = \frac{11!}{3!(11-3)!} = \frac{11!}{3!8!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165$ .
Multiply Total Combinations: Next, we calculate the number of ways to choose $5$ students from $12$ . Using the combination formula again, we get $C(12, 5) = \frac{12!}{5!(12-5)!} = \frac{12!}{5!7!} = \frac{(12 \times 11 \times 10 \times 9 \times 8)}{(5 \times 4 \times 3 \times 2 \times 1)} = 792$ .
Multiply Total Combinations: Next, we calculate the number of ways to choose $5$ students from $12$ . Using the combination formula again, we get $C(12, 5) = \frac{12!}{5!(12-5)!} = \frac{12!}{5!7!} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792$ .Now, we multiply the number of combinations of teachers by the number of combinations of students to find the total number of ways to form the committee. So, we have $165$ (ways to choose teachers) $\times 792$ (ways to choose students) $= 130680$ .

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