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A committee must be formed with 4 teachers and 4 students. If there are 7 teachers to choose from, and 9 students, how many different ways could the committee be made?
Answer:

A committee must be formed with $4$ teachers and $4$ students. If there are $7$ 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

4

teachers and

4

students. If there are

7

teachers to choose from, and

9

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 $4$ 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 number of ways to choose $4$ teachers out of $7$ . Using the combination formula: $\newline$ $C(7, 4) = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{(7\times6\times5\times4!)}{(4!\times3\times2\times1)} = \frac{(7\times6\times5)}{(3\times2\times1)} = 35$ . $\newline$ There are $35$ different ways to choose the teachers.
Calculate Total Ways: Next, we calculate the number of ways to choose $4$ students out of $9$ . Again, using the combination formula: $\newline$ $C(9, 4) = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = \frac{9\times8\times7\times6\times5!}{4!\times5!} = \frac{9\times8\times7\times6}{4\times3\times2\times1} = 126$ . $\newline$ There are $126$ different ways to choose the students.
Calculate Total Ways: Next, we calculate the number of ways to choose $4$ students out of $9$ . Again, using the combination formula: $\newline$ $C(9, 4) = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = \frac{9\times8\times7\times6\times5!}{4!\times5!} = \frac{9\times8\times7\times6}{4\times3\times2\times1} = 126$ . $\newline$ There are $126$ different ways to choose the students.Finally, to find the total number of different ways to form the committee, we multiply the number of ways to choose the teachers by the number of ways to choose the students: $\newline$ Total ways = Number of ways to choose teachers $\times$ Number of ways to choose students = $35 \times 126 = 4410$ .

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