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

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Senior Manager
Joined: 23 May 2005
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Anthony and Michael sit on the six-member board of directors [#permalink]  01 Nov 2006, 08:56
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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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%
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Impossible is nothing

Manager
Joined: 01 Nov 2006
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There are 6C3 = 20 possible 3 person subcommittees. The number including M&A = 4 so it is 4/20 = 20% as Peter wrote.
Intern
Joined: 02 Aug 2006
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Could that be C- 40%? It seems to be a word problem as well.

The question says "what percent of all possible committees that include Michael also include Anthony". Out of 20 committees obviously only half include Michael, thus 4 out of 10 - 40%.

Please let me know if I am wrong.
Senior Manager
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Location: Phoenix, AZ, USA
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20% [#permalink]  02 Nov 2006, 08:13
Total number of comeetee possible

6C3= 6!/3!*3!= 6*5*4/3*2=20

For number of comette when both michal and antony are togather

we know two members have been pciked

so we have 4 people on 1 person to be chosen to make it cometee of 3

so 4C1=4!/3!=4

% will be 4/20= 20%
Manager
Joined: 29 Jul 2006
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One more for C, there are 8 subcommittes that can be formed with Anthony and Michael:
4 with Anthony and Micheal in the first subcommittee
4 with Anthony and Micheal in the second subcommittee
VP
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Re: PS: Perm/Comb problem [#permalink]  02 Nov 2006, 11:12
Hermione 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%

A is right.

6C3 combinations yield 20 possibilties for 3 member groups.

Constraining 2 members to be together.. we are left with 4 other members to be chosen 1 at a time since the total subgroup strength is 3. There are 4 ways of choosing 4 members and pairing them with the 2 fixed members.

So 4 groups will have 2 fixed members.
Senior Manager
Joined: 23 May 2005
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Location: Sing/ HK
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Guys, the correct answer is C. 40%

Question is asking about all the possible subcommittees that include Michael. And from that, what percent of these subcommittees also include Anthony.

5!/ 2!3! = 10. This is the number of committees where Michael is a member.

Now how many of these committees will Anthony also be a member? Since there are only 4 people left, then 4.

We get 4/10 = 40%. Of the 10 committees that include Michael, 4 include anthony.

Voila
_________________

Impossible is nothing

Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK
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Kudos [?]: 19 [0], given: 0

I had a tough time solving this one... question stem is confusing...
_________________

Impossible is nothing

Intern
Joined: 02 Aug 2006
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Location: NYC
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Hermione wrote:
Guys, the correct answer is C. 40%

Question is asking about all the possible subcommittees that include Michael. And from that, what percent of these subcommittees also include Anthony.

5!/ 2!3! = 10. This is the number of committees where Michael is a member.

Now how many of these committees will Anthony also be a member? Since there are only 4 people left, then 4.

We get 4/10 = 40%. Of the 10 committees that include Michael, 4 include anthony.

Voila

Intern
Joined: 29 Dec 2005
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The question stem says 2 committees of 3 persons each.

Ans is A (20%)
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