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A committee must be formed with 4 teachers and 3 students. If there are 10 teachers to choose from, and 16 students, how many different ways could the committee be made?
Answer:

A committee must be formed with $4$ teachers and $3$ students. If there are $10$ teachers to choose from, and $16$ 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

4

teachers and

3

students. If there are

10

teachers to choose from, and

16

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 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 Combination: First, calculate the number of ways to choose $4$ teachers out of $10$ . Using the combination formula: $\newline$ $C(10, 4) = \frac{10!}{4!(10 - 4)!} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$ ways to choose the teachers.
Multiply Total Combinations: Next, calculate the number of ways to choose $3$ students out of $16$ . Using the combination formula: $\newline$ $C(16, 3) = \frac{16!}{3!(16 - 3)!} = \frac{16!}{3!13!} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560$ ways to choose the students.
Multiply Total Combinations: Next, calculate the number of ways to choose $3$ students out of $16$ . Using the combination formula: $\newline$ $C(16, 3) = \frac{16!}{3!(16 - 3)!} = \frac{16!}{3!13!} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560$ ways to choose the students.Now, multiply the number of ways to choose teachers by the number of ways to choose students to find the total number of different committees that can be formed: $\newline$ Total number of committees = $210$ (ways to choose teachers) $\times 560$ (ways to choose students) = $117,600$ different committees.

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