-
Stockmoose16
- Master | Next Rank: 500 Posts
- Posts: 347
- Joined: Mon Aug 04, 2008 1:42 pm
- Thanked: 1 times
Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?
20%
30%
40%
50%
60%
This question has been posted before, but I'm wondering if the following method is a viable way to get the answer:
Since Michael and Anthony need to be on the same committee, fix their position:
MA4
The "4" stands for the remaining people who can fill the seat on a committee with Michael and Anthony. Since the committee that contains M and A can be arranged in any manner, you must multiply by 3!. So there are 3! * 4= 24 ways to organize each committee. There are 2 possible committees = 48.
Each committee can be arranged in 6*5*4 ways =120
So the answer is: 48/120= 40%
Is this a viable way to get the answer? I'm wondering if multiplying by 3! is INCORRECT, since a committee with MXA is the same as XMA, thus, we'd be double counting.
Can an expert weigh in?
20%
30%
40%
50%
60%
This question has been posted before, but I'm wondering if the following method is a viable way to get the answer:
Since Michael and Anthony need to be on the same committee, fix their position:
MA4
The "4" stands for the remaining people who can fill the seat on a committee with Michael and Anthony. Since the committee that contains M and A can be arranged in any manner, you must multiply by 3!. So there are 3! * 4= 24 ways to organize each committee. There are 2 possible committees = 48.
Each committee can be arranged in 6*5*4 ways =120
So the answer is: 48/120= 40%
Is this a viable way to get the answer? I'm wondering if multiplying by 3! is INCORRECT, since a committee with MXA is the same as XMA, thus, we'd be double counting.
Can an expert weigh in?
