I am trying to work this problem out. I know that the systematic counting principle is used. However, I cannot get the right answer.
For the first part, I took
R=2
N=6
6!/(6-2)! (Permutation).
Answer: 30 choices for the Chair/Vice
The second part, I took
R=2
N=8
Combination form.
From here, I do not know what to do or how to get the correct answer. Please explain how to get the answer. Thank You!
The academic computing committee at a college is in the process of evaluating different computer systems. The committee consists of six administrators, six faculty, and two students. A six-person subcommittee is to be formed. The subcommittee must have a chair and vice chair from the administrators, the other four committee members have no particularly defined roles from the faculty and students. In how many ways can this subcommittee be formed?
2 out of 6 administrators
4 out of 6 others
combinations of 6 taken 2 at a time
= 6!/[ 2!(4!)]
=6*5/2
= 15 combinations of administrators
combinations of 6 taken 4 at a time
= 6!/[ 4!(2!)] = 6*5/2 = 15 combinations of faculty and students
so 15*15 = 225
the other 4 committee members come from the group of 6 faculty and 2 students ... 8C4
number of ways is ... 6P2 * 8C4
8C4 = 8!/[4!(4!)]
= 8*7*6*5/(4*3*2)
= 70
so 15*70 = 1050
Thank You all for your feedback!
I figured out how to solve this problem and the answer was 2100.
To solve this problem, we need to break it down into two parts: selecting the chair and vice chair from the administrators, and selecting the remaining four members from the faculty and students.
Part 1: Selecting the chair and vice chair from the administrators
Since there are 6 administrators and we need to select 2 of them, we can use the permutation formula. The formula for permutations is given by:
P(n, r) = n! / (n - r)!
In this case, we have n = 6 (number of administrators) and r = 2 (number of positions to fill - chair and vice chair). Plugging the values into the formula, we get:
P(6, 2) = 6! / (6 - 2)! = 6! / 4! = (6 * 5 * 4 * 3 * 2 * 1) / (4 * 3 * 2 * 1) = 30
So there are 30 ways to select the chair and vice chair from the administrators.
Part 2: Selecting the remaining four members from the faculty and students
Since the remaining four members have no particularly defined roles, we can use combinations to select them. The formula for combinations is given by:
C(n, r) = n! / ((n - r)! * r!)
In this case, we have n = 8 (number of faculty and students combined) and r = 4 (number of positions to fill). Plugging the values into the formula, we get:
C(8, 4) = 8! / ((8 - 4)! * 4!) = (8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / ((4 * 3 * 2 * 1) * (4 * 3 * 2 * 1)) = 70
So there are 70 ways to select the remaining four members from the faculty and students.
To find the total number of ways to form the subcommittee, we multiply the number of options for each part:
Total = Part 1 * Part 2
Total = 30 * 70 = 2100
Therefore, there are 2100 ways to form the subcommittee in this scenario.