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A committee that includes 6 members is about to be divided

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Manager
Joined: 14 Jul 2014
Posts: 192

Kudos [?]: 18 [0], given: 110

Location: United States
GMAT 1: 600 Q48 V27
GMAT 2: 720 Q50 V37
GPA: 3.2
Re: A committee that includes 6 members is about to be divided [#permalink]

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19 Jan 2016, 11:16
If first team has Michael and David, there will be 4 ways to arrange that team. M, D, _ . (Last member can be one of the remaining four).
Second team will be arranged only in one way, order does not matter. So 4 * 1 = 4 ways for first team having M and D.

Repeat for second team = 4 ways.

Total number of possibilities: 6C3 = 20.

So, 8 possible ways out of 20 = 40 percent.

Kudos [?]: 18 [0], given: 110

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Re: A committee that includes 6 members is about to be divided [#permalink]

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22 Feb 2017, 06:44
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Re: A committee that includes 6 members is about to be divided [#permalink]

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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

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Re: A committee that includes 6 members is about to be divided [#permalink]

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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%

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Re: A committee that includes 6 members is about to be divided [#permalink]

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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

Kudos [?]: 23 [0], given: 562

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

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