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Anthony and Michael sit on the six-member board of directors [#permalink]

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05 Feb 2006, 22:35

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Question Stats:

36% (02:35) correct
64% (01:31) wrong based on 95 sessions

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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?

Re: Anthony and Michael sit on the six-member board of directors [#permalink]

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05 Feb 2006, 22:56

Is it 40%?

Total possibilities = 6C3 = 20

A & M can be in one in committee in 4 ways. (A,M,X where X can be any of the other 4 members)

There are 2 such committees. Hence total num of committes in which A &M are together = 4*2 = 8

Ans = 8/20 * 100 = 40% _________________

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Re: Anthony and Michael sit on the six-member board of directors [#permalink]

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06 Feb 2006, 06:27

all possible SUB COMS THAT INCLUDE M are 5C2=10 or 10 options.
When M and A are together then we have 4C1 or 4 options .
Then 4/10 of all subcommittees that include M also include A
So i get 40%
may be i am wrong

"# of possible to form 3 person team from 6 people = 6C3 = 20. That leaves another 20 combinations for the other team. Total = 40 possibilites (20 team A combination, 20 team B combination).

If Micheal is in team A, then Anthony has to be in team A, and team A will have only 4 ways to form (due to 4 people remaining).

But Micheal and Anthony can be in team A or team B. So % = (4/20)*2*100 = 40%"

So, we must multiply with 2!

Last edited by allabout on 06 Feb 2006, 08:08, edited 1 time in total.

Re: Anthony and Michael sit on the six-member board of directors [#permalink]

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06 Feb 2006, 08:07

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Total possible combination = 6C3 = 20

Now let the members be A B C D E M
total number of 3-member-committe in which A and M are present= 4 [NOT 8]
because
(AMB , AMC, AMD, AME) is same as (MAB, MAC, MAD, MAE)

Re: Anthony and Michael sit on the six-member board of directors [#permalink]

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06 Feb 2006, 08:55

BG wrote:

all possible SUB COMS THAT INCLUDE M are 5C2=10 or 10 options. When M and A are together then we have 4C1 or 4 options . Then 4/10 of all subcommittees that include M also include A So i get 40% may be i am wrong

Re: Anthony and Michael sit on the six-member board of directors [#permalink]

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08 Jun 2015, 06:48

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Re: Anthony and Michael sit on the six-member board of directors [#permalink]

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08 Jun 2015, 06:49

mxr55820 wrote:

Total possible combination = 6C3 = 20

Now let the members be A B C D E M total number of 3-member-committe in which A and M are present= 4 [NOT 8] because (AMB , AMC, AMD, AME) is same as (MAB, MAC, MAD, MAE)

(4/20)*100 = 20 answer

total possible combinations = 6C3 x 2 = 40 because 2 subcom have to be formed

Re: Anthony and Michael sit on the six-member board of directors [#permalink]

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08 Jun 2015, 06:53

Expert's post

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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? A. 20% B. 30% C. 40% D. 50% E. 60%

First approach: Let's take the group with Michael: there is a place for two other members and one of them should be taken by Anthony, as there are total of 5 people left, hence there is probability of 2/5=40%.

Second approach: Again in Michael's group 2 places are left, # of selections of 2 out of 5 5C2=10 - total # of outcomes. Select Anthony - 1C1=1, select any third member out of 4 - 4C1=4, total # =1C1*4C1=4 - total # of winning outcomes. P=# of winning outcomes/# of outcomes=4/10=40%

Third approach: Michael's group: Select Anthony as a second member out of 5 - 1/5 and any other as a third one out of 4 left 4/4, total=1/5*4/4=1/5; Select any member but Anthony as second member out of 5 - 4/5 and Anthony as a third out of 4 left 1/4, total=4/5*1/4=1/5; Sum=1/5+1/5=2/5=40%

Fourth approach: Total # of splitting group of 6 into two groups of 3: 6C3*3C3/2!=10 # of groups with Michael and Anthony: 1C1*1C1*4C1=4 P=4/10=40%

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