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?
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 committee members.
Part 1: Selecting the Chair and Vice Chair from the Administrators
In this part, we need to select 2 administrators to be the chair and vice chair of the subcommittee.
We will use the permutation formula, since the order matters (chair and vice chair have different roles).
We have:
R = 2 (we need to select 2 administrators)
N = 6 (we have 6 administrators)
Using the permutation formula, the number of ways to select the chair and vice chair is:
6P2 = 6! / (6-2)! = 6! / 4! = 6 * 5 = 30 choices for the chair and vice chair.
Part 2: Selecting the Remaining Committee Members
In this part, we need to select 4 members from the remaining administrators, faculty, and students (since the chair and vice chair were already selected). The order doesn't matter here.
We will use the combination formula, since the order doesn't matter.
We have:
R = 4 (we need to select 4 members)
N = 6 administrators + 6 faculty + 2 students = 14 (total number of remaining members)
Using the combination formula, the number of ways to select the remaining committee members is:
14C4 = 14! / (4!(14-4)!) = 14! / (4! * 10!) = (14 * 13 * 12 * 11) / (4 * 3 * 2 * 1) = 1001 choices for the remaining committee members.
To get the total number of ways to form the subcommittee, we multiply the number of choices for the chair and vice chair by the number of choices for the remaining committee members:
Total number of ways = 30 * 1001 = 30,030 ways to form the subcommittee.
Therefore, there are 30,030 ways to form the subcommittee.