Problem Solving

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?

I think your method calculates M&A/All rather than M&A/M. The combos for Michael should be in the denominator rather than the total combos.

M&A/M = 4/5C2 = 40%

pseudononymous wrote:I think your method calculates M&A/All rather than M&A/M. The combos for Michael should be in the denominator rather than the total combos.

M&A/M = 4/5C2 = 40%

I think my denominator only calculates the scenarios where Michael and Anthony are on the same committee. Total outcomes without regard to the people would be 6!.

pseudononymous wrote:I think your method calculates M&A/All rather than M&A/M. The combos for Michael should be in the denominator rather than the total combos.

M&A/M = 4/5C2 = 40%

Also, how did you get M&A being on the same committee as "4"?

It can be XMA, XAM, AMX, AXM, MAX, MXA... that's 6*2 committees =12

Stockmoose16 wrote:

pseudononymous wrote:I think your method calculates M&A/All rather than M&A/M. The combos for Michael should be in the denominator rather than the total combos.

M&A/M = 4/5C2 = 40%

Also, how did you get M&A being on the same committee as "4"?

It can be XMA, XAM, AMX, AXM, MAX, MXA... that's 6*2 committees =12

i was talking about unique combos where order doesn't matter

pseudononymous wrote:

Stockmoose16 wrote:

pseudononymous wrote:I think your method calculates M&A/All rather than M&A/M. The combos for Michael should be in the denominator rather than the total combos.

M&A/M = 4/5C2 = 40%

Also, how did you get M&A being on the same committee as "4"?

It can be XMA, XAM, AMX, AXM, MAX, MXA... that's 6*2 committees =12

i was talking about unique combos where order doesn't matter

But order does matter when you're doing probability. And this is a probability question. If I say, "what are the chances of picking a red marble and a blue marble from a bag that contains 4 red marbles and 6 blue marbles, you need to calculate the chances of getting a red then a blue, and then the other way around (blue then red).

Why do you think order doesn't matter?

In this problem, MAX is the same committee as XAM which is why order doesn't matter.

Numerator (4C1=4):
MA1
MA2
MA3
MA4

Denominator (5C2=10):
M12
M13
M14
M15
M23
M24
M25
M34
M35
M45

4/10 = 40%

If you care about order, you can multiply both the numerator and denominator by 3! which will give you the same answer:

4C1*3! / 5C2*3! = 24 / 60 = 40%