Answer:
292,600 committees are possible.
Step-by-step explanation:
The order is not important.
That means that, for example, a comittee of students Tre'davious and Tremaine is the same comittee as Tremaine and Tre'davious. So the combinations formula is used to solve this problem
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In this problem, we have that:
Initially, we have to find the number of groups of administrators, faculty members and students separated. Then we multiply these values to find how many six-member committees are possible.
Administrators
There are 7 administrators in contention. 1 is selected.
So there are 7 possible ways to select an administrator.
Faculty members
There are 12 faculty members in contention. 3 are selected.
[tex]C_{12,3} = \frac{12!}{3!9!} = 220[/tex]
So there are 220 ways to select groups of 3 faculty members.
Students
There are 20 students in contention. 2 are selected.
[tex]C_{20,2} = \frac{20!}{2!18!} = 190[/tex]
So there are 190 ways to select groups of 2 students.
How many six-member committees are possible?
For each administrator, there are 220 groups of faculty members.
For each group of faculty members, there are 190 groups of students.
There are 7 possible administrators
7*220*190 = 292,600
292,600 committees are possible.