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A committee must be formed with 5 teachers and 3 students. If there are 7 teachers to choose from, and 8 students, how many different ways could the committee be made?
Answer:

A committee must be formed with $5$ teachers and $3$ students. If there are $7$ teachers to choose from, and $8$ 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

5

teachers and

3

students. If there are

7

teachers to choose from, and

8

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 need to choose $5$ out of $7$ , which is a combination problem. The formula for combinations 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 combinations of teachers. Using the formula, we have $C(7, 5) = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!} = \frac{(7\times6)}{(2\times1)} = \frac{42}{2} = 21$ . There are $21$ different ways to choose $5$ teachers from $7$ .
Multiply Combinations: Next, we calculate the combinations of students. We need to choose $3$ out of $8$ , which is another combination problem. Using the formula, we have $C(8, 3) = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{(8\times7\times6)}{(3\times2\times1)} = \frac{336}{6} = 56$ . There are $56$ different ways to choose $3$ students from $8$ .
Multiply Combinations: Next, we calculate the combinations of students. We need to choose $3$ out of $8$ , which is another combination problem. Using the formula, we have $C(8, 3) = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{(8\times7\times6)}{(3\times2\times1)} = \frac{336}{6} = 56$ . There are $56$ different ways to choose $3$ students from $8$ .Finally, we multiply the number of combinations of teachers by the number of combinations of students to find the total number of different ways to form the committee. So, we have $21$ (ways to choose teachers) $\times$ $56$ (ways to choose students) $= 1176$ .

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