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A committee must be formed with 3 teachers and 6 students. If there are 11 teachers to choose from, and 16 students, how many different ways could the committee be made?
Answer:

A committee must be formed with $3$ teachers and $6$ students. If there are $11$ 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

3

teachers and

6

students. If there are

11

teachers to choose from, and

16

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

\newline

Answer:

Choose Teachers Calculation: Determine the number of ways to choose $3$ teachers out of $11$ . We 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. For the teachers, $n = 11$ and $k = 3$ . Calculate the number of combinations for teachers: $C(11, 3) = \frac{11!}{3!(11-3)!} = \frac{11!}{3!8!} = \frac{11\times10\times9}{3\times2\times1} = 165$ .
Choose Students Calculation: Determine the number of ways to choose $6$ students out of $16$ . Again, we use the combination formula. For the students, $n = 16$ and $k = 6$ . Calculate the number of combinations for students: $C(16, 6) = \frac{16!}{(6!(16-6)!)} = \frac{16!}{(6!10!)} = \frac{(16\times15\times14\times13\times12\times11)}{(6\times5\times4\times3\times2\times1)} = 8008$ .
Total Number of Ways Calculation: Calculate the total number of ways to form the committee by multiplying the number of combinations of teachers by the number of combinations of students. $\newline$ Total number of ways = Number of ways to choose teachers $\times$ Number of ways to choose students = $165 \times 8008$ . $\newline$ Perform the multiplication: $165 \times 8008 = 1321320$ .

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