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A committee must be formed with 4 teachers and 6 students. If there are 6 teachers to choose from, and 14 students, how many different ways could the committee be made?
Answer:

A committee must be formed with $4$ teachers and $6$ students. If there are $6$ teachers to choose from, and $14$ 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

6

students. If there are

6

teachers to choose from, and

14

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

\newline

Answer:

Calculate Teachers Combination: We need to calculate the number of ways to choose $4$ teachers out of $6$ and $6$ students out of $14$ . This is a combination problem where order does not matter. 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, we calculate the number of ways to choose $4$ teachers out of $6$ . Using the combination formula, we get $C(6, 4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!} = \frac{(6\times5\times4\times3\times2\times1)}{((4\times3\times2\times1)(2\times1))} = \frac{(6\times5)}{(2\times1)} = 15$ .
Multiply Total Combinations: Next, we calculate the number of ways to choose $6$ students out of $14$ . Using the combination formula, we get $C(14, 6) = \frac{14!}{6!(14-6)!} = \frac{14!}{6!8!} = \frac{14\times13\times12\times11\times10\times9\times8!}{6!8!} = \frac{14\times13\times12\times11\times10\times9}{6\times5\times4\times3\times2\times1} = 3003$ .
Multiply Total Combinations: Next, we calculate the number of ways to choose $6$ students out of $14$ . Using the combination formula, we get $C(14, 6) = \frac{14!}{(6!(14-6)!)} = \frac{14!}{(6!8!)} = \frac{(14\times13\times12\times11\times10\times9\times8!)}{(6!8!)} = \frac{(14\times13\times12\times11\times10\times9)}{(6\times5\times4\times3\times2\times1)} = 3003$ . 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. So, we have $15$ (ways to choose teachers) $\times 3003$ (ways to choose students) $= 45045$ .

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