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A committee must be formed with 5 teachers and 4 students. If there are 8 teachers to choose from, and 15 students, how many different ways could the committee be made?
Answer:

A committee must be formed with $5$ teachers and $4$ students. If there are $8$ teachers to choose from, and $15$ 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

4

students. If there are

8

teachers to choose from, and

15

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

\newline

Answer:

Calculate Teachers Combination: 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 $8$ , 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 Combination: First, we calculate the number of ways to choose $5$ teachers out of $8$ . Using the combination formula: $\newline$ $C(8, 5) = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!} = \frac{(8\times7\times6)}{(3\times2\times1)} = 56$ ways to choose the teachers.
Multiply Teachers and Students: Next, we calculate the number of ways to choose $4$ students out of $15$ . Again, using the combination formula: $\newline$ $C(15, 4) = \frac{15!}{4!(15-4)!} = \frac{15!}{4!11!} = \frac{(15\times14\times13\times12)}{(4\times3\times2\times1)} = 1365$ ways to choose the students.
Total Number of Committees: Now, we multiply the number of ways to choose the teachers by the number of ways to choose the students to find the total number of different committees that can be formed. $\newline$ Total number of committees $=$ Number of ways to choose teachers $\times$ Number of ways to choose students $= 56 \times 1365$ .
Perform Multiplication: Performing the multiplication gives us the total number of different committees: $\newline$ Total number of committees = $56 \times 1365 = 76440$ .

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