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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] New post 01 Oct 2010, 07:39
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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%
[Reveal] Spoiler: OA
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Re: Probability - MGMAT Test [#permalink] New post 01 Oct 2010, 07:42
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Barkatis wrote:
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%

Answer: C.

Hope it helps.
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Re: Probability - MGMAT Test [#permalink] New post 01 Oct 2010, 08:05
Thanks.
Just a question: I noticed that you responded very quickly to my posts. Which I thank you for. But I am wondering if it's ok that I post questions that have been obviously posted before. If it's a problem then please tell me how can I check whether a question is already on the forum or not. Thanks.
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Re: Probability - MGMAT Test [#permalink] New post 01 Oct 2010, 08:26
Expert's post
Barkatis wrote:
Thanks.
Just a question: I noticed that you responded very quickly to my posts. Which I thank you for. But I am wondering if it's ok that I post questions that have been obviously posted before. If it's a problem then please tell me how can I check whether a question is already on the forum or not. Thanks.


Generally it's a good idea to do the search before posting, for example I would search in PS subforum (there is a search field above the topics) for the word "Anthony", don't think that there are many questions with this name. Though it's not a probelm to post a question that was posted before: if moderators find previous discussions they will merge the topics, copy the solution from there or give a link to it.
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Re: Probability - MGMAT Test [#permalink] New post 04 Oct 2010, 11:56
Bunuel, A question just came to my mind concerning the second approach of the solution that you proposed:

total # of winning outcomes
Select Anthony: 1C1=1, select any third member out of 4: 4C1=4, total # =1C1*4C1=4

If we say 1C1*4C1 don't we assume that the winning can be either (Anthony,Third member) or (Third member, Anthony) but not both ?
shouldn't we multiply 1C1*4C1 by 2 to get the total number of possible combinations ?
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Re: Probability - MGMAT Test [#permalink] New post 04 Oct 2010, 12:12
Expert's post
Barkatis wrote:
Bunuel, A question just came to my mind concerning the second approach of the solution that you proposed:

total # of winning outcomes
Select Anthony: 1C1=1, select any third member out of 4: 4C1=4, total # =1C1*4C1=4

If we say 1C1*4C1 don't we assume that the winning can be either (Anthony,Third member) or (Third member, Anthony) but not both ?
shouldn't we multiply 1C1*4C1 by 2 to get the total number of possible combinations ?


We are counting # of committees with Anthony and Michael:

{M,A,1};
{M,A,2};
{M,A,3};
{M,A,4}.

Here {M,A,1} is the same committee as {M,1,A}.
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Re: Probability - MGMAT Test [#permalink] New post 21 Jan 2011, 15:54
Bunuel, can you advise why do we have to divide 6C3 * 3C3 by 2!, in the fourth approach ?
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Re: Probability - MGMAT Test [#permalink] New post 21 Jan 2011, 16:14
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praveenvino wrote:
Bunuel, can you advise why do we have to divide 6C3 * 3C3 by 2!, in the fourth approach ?


We are dividing by 2! (factorial of the # of groups) because the order of the groups is not important (we don't have the group #1 and the group #2) and we need to get rid of the duplications.

Dividing a group into subgroups:
combinations-problems-95344.html
split-the-group-101813.html
9-people-and-combinatorics-101722.html
ways-to-divide-99053.html
combination-and-selection-into-team-106277.html
ways-to-split-a-group-of-6-boys-into-two-groups-of-3-boys-ea-105381.html
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COLLECTION OF QUESTIONS:
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Re: Anthony and Michael sit on the six-member board of directors [#permalink] New post 28 Dec 2012, 01:39
Barkatis wrote:
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%


Here is my approach:

I first counted the possible creation of 2 subcommittees without restriction: \frac{6!}{3!3!}* \frac{3!}{3!}= 20
Then, I now proceed to counting the number of ways to create committee with Michael and Anthony together.

M A _ + _ _ _ = \frac{4!}{1!3!} * \frac{1!}{1!} = 4

MA could be in group#1 or group#2. Thus, =4*2 = 8

Final calculation: 8/20 = 4/10 = 40%

Answer: C
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Re: Anthony and Michael sit on the six-member board of directors [#permalink] New post 20 Jan 2014, 09:45
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Re: Anthony and Michael sit on the six-member board of directors   [#permalink] 20 Jan 2014, 09:45
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