A school has $8$ sixth-grade teachers and $7$ administrators. In order to go on a field trip, $2$ teachers and $1$ administrator need to accompany the students. How many different groups of adults can go on the field trip?

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Identify equivalent linear expressions: word problems

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Q. A school has

8

sixth-grade teachers and

7

administrators. In order to go on a field trip,

2

teachers and

1

administrator need to accompany the students. How many different groups of adults can go on the field trip?

Calculate Teachers Combination: To find the number of different groups of adults that can go on the field trip, we need to calculate the number of ways to choose $2$ teachers out of $8$ and $1$ administrator out of $7$ . This is a combination problem where order does not matter.
Calculate Administrators Combination: First, calculate the number of ways to choose $2$ teachers out of $8$ . This is given by the combination formula $C(n, k) = \frac{n!}{k!(n-k)!}$ , where $n$ is the total number and $k$ is the number chosen. So, $C(8, 2) = \frac{8!}{(2!(8-2)!)} = \frac{8!}{(2!6!)}$ .
Multiply Teachers and Administrators: Perform the calculation for $C(8, 2)$ . We have $\frac{8!}{(2!6!)} = \frac{(8\times7\times6!)}{(2\times1\times6!)} = \frac{(8\times7)}{(2\times1)} = 28$ . So there are $28$ ways to choose $2$ teachers.
Incorrect Approach Explanation: Next, calculate the number of ways to choose $1$ administrator out of $7$ . This is given by $C(7, 1) = \frac{7!}{(1!(7-1)!)} = \frac{7!}{(1!6!)} = 7$ .
Incorrect Approach Explanation: Next, calculate the number of ways to choose $1$ administrator out of $7$ . This is given by $C(7, 1) = \frac{7!}{(1!(7-1)!)} = \frac{7!}{(1!6!)} = 7$ .Since the selection of teachers and administrators are independent events, we multiply the number of ways to choose the teachers by the number of ways to choose the administrator. So, the total number of different groups is $28$ (from the teachers) $\times 7$ (from the administrators) $= 196$ .
Incorrect Approach Explanation: Next, calculate the number of ways to choose $1$ administrator out of $7$ . This is given by $C(7, 1) = \frac{7!}{1!(7-1)!} = \frac{7!}{1!6!} = 7$ .Since the selection of teachers and administrators are independent events, we multiply the number of ways to choose the teachers by the number of ways to choose the administrator. So, the total number of different groups is $28$ (from the teachers) $\times 7$ (from the administrators) $= 196$ .The given expression $\left(\frac{c(8,2)\times c(1,1)}{c(15,3)}\right)\left(\frac{(8!)}{(2!6!)}\times\frac{(7!)}{(6!)}\right)/\left(\frac{(15!)}{(3!12!)}\right)\times\frac{(4\times7\times7)}{(5\times14\times13)}$ seems to be an incorrect approach to the problem and includes unnecessary calculations. The correct approach is simply to multiply the combinations of teachers and administrators as calculated in the previous steps.