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Re: A committee that includes 6 members is about to be divided
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16 Apr 2017, 12:37

sstasic wrote:

This is equivalent to: We have 6 chairs divided in 2 Groups: A and B each with 3 chairs. If Michael sits on the first chair in the group A, what is the posibility that David sits on any of another 2 available chairs in Group A?

So: A: M _ _ / B: _ _ _ There are 2 chairs in Group A for favorable outcomes and 5 available chairs as total possible outcomes. The answer is 2/5

This is the only explanation that 100% makes sense to me. Kudos to you

Re: A committee that includes 6 members is about to be divided
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16 Apr 2017, 13:42

M and D in different groups: (2/6)*(4/5)*(3/4)*3*2 = 6/10 = 60% (2/6: choose M or D from 6 people 4/5: 4 ways to choose 1 person from 5 people, except M or D who has been already chosen 3/4: 3 ways to choose the last person for the 1st group 3 different orders of choosing the 1st 3 persons for the 1st group 2 different orders of choosing the groups) --> M and D in the same group: 1 - 60% = 40%

Re: A committee that includes 6 members is about to be divided
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16 Apr 2017, 15:10

now, there is a similar problem, if there are 9 committees divided into 3 equal sub-committees, what is chance that both Michael and Bose are in the same sub?

According to re-interpretation of the problem, it should be 2/8 = 25% My method, also known as traditional method, is 7C1 / 9C3 = 1/12 for the first sub, then the chance should be 3 * 1/12 = 1/4

Re: A committee that includes 6 members is about to be divided
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22 Apr 2018, 18:07

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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