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I am trying to work this problem out. I know that the systematic counting principle is used. However, I cannot get the right an...Asked by Johnathan (John)
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?
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?
Answers
Answered by
Damon
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
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
Answered by
Scott
the other 4 committee members come from the group of 6 faculty and 2 students ... 8C4
number of ways is ... 6P2 * 8C4
number of ways is ... 6P2 * 8C4
Answered by
Damon
8C4 = 8!/[4!(4!)]
= 8*7*6*5/(4*3*2)
= 70
so 15*70 = 1050
= 8*7*6*5/(4*3*2)
= 70
so 15*70 = 1050
Answered by
Johnathan (John)
Thank You all for your feedback!
I figured out how to solve this problem and the answer was 2100.
I figured out how to solve this problem and the answer was 2100.
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