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

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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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95% (hard)

Question Stats:

39% (01:35) correct 61% (01:31) wrong based on 119 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?

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: anthony-and-michael-sit-on-the-six-member-board-of-directors-102027.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Jun 2015, 06:53, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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

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08 Jun 2015, 06:53
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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%

OPEN DISCUSSION OF THIS QUESTION IS HERE: anthony-and-michael-sit-on-the-six-member-board-of-directors-102027.html
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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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05 Feb 2006, 23:00
, i dont know how i thot 20% , but giddi is my hero and i am learning from him.. so i take wht u say sir!
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Re: Anthony and Michael sit on the six-member board of directors [#permalink]

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06 Feb 2006, 00:38
# of 3 people teams = 6C3 = 20

# of 3 people teams with Michael and Anthony = M+A+any remaining 4 -> 4 ways

% = 4/20 * 100% = 20%

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

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06 Feb 2006, 00:40
giddi77 wrote:
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%

I think you do not need to multiply by 2 as we're concerned with who's in the team, and not what position they hold in the team.

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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

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

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06 Feb 2006, 07:44
number of ways that the two persons are included= 4C1=4

total number of ways if choose randomly = 6C3 =20

the percentage = 4/20 = 20%

Last edited by celiaXDN on 06 Feb 2006, 07:51, edited 2 times in total.

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

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06 Feb 2006, 08:06
Ywilfried the last time you did this question your answer was 40% and mine was 20%, remember it? You gave a nice explanation:

http://www.gmatclub.com/phpbb/viewtopic.php?t=24644

"# 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.

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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)

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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

BG,

That is nice and quick way explanation..

OA is 40%

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

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08 Jun 2015, 06:48
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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)

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

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Re: Anthony and Michael sit on the six-member board of directors   [#permalink] 08 Jun 2015, 06:49
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