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A committee must be formed with 4 teachers and 3 students. If there are 9 teachers to choose from, and 13 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 $9$ teachers to choose from, and $13$ 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

9

teachers to choose from, and

13

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

\newline

Answer:

Calculate Teachers Combination: Calculate the number of ways to choose $4$ teachers out of $9$ . 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 teachers, $n = 9$ and $k = 4$ . $C(9, 4) = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = \frac{9\times8\times7\times6}{4\times3\times2\times1} = 126$
Calculate Students Combination: Calculate the number of ways to choose $3$ students out of $13$ . Again, we use the combination formula. For students, $n = 13$ and $k = 3$ . $C(13, 3) = \frac{13!}{3!(13-3)!} = \frac{13!}{3!10!} = \frac{13\times12\times11}{3\times2\times1} = 286$
Calculate Total Number of Ways: Calculate the total number of ways to form the committee by multiplying the number of ways to choose teachers by the number of ways to choose students. $\newline$ Total number of ways $=$ Number of ways to choose teachers $\times$ Number of ways to choose students $\newline$ Total number of ways $= 126 \times 286$ $\newline$ Total number of ways $= 36036$

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