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

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Director
Joined: 10 Feb 2006
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Anthony and Michael sit on the six-member board of directors [#permalink]  28 Sep 2006, 11:45
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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Manager
Joined: 28 Aug 2006
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Location: Albuquerque, NM
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total ways to choose the committees = 6C3 = 20, since once you choose one other one is already defined

no. of ways in which A and M are together = 4 (once u choose two of them only four choices remain for the 3rd person)

percentage = 4/20 = 20%
Manager
Joined: 25 May 2006
Posts: 227
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Getting 40%

Assume we have the following members
M-A-B-C-D-E

Possible combinations that with M also include A:
MAB
MAC
MAE

MBC
MBD
MBE
MCD
MCE
MDE

TOTAL of 10 Combinations. 4 out of 10 includes A.
So, 4/10=40%
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Senior Manager
Joined: 28 Aug 2006
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It is C.

Now the possible committees that include M :

Totally we need 3 people out which M is already there. So the remaining two could be any 2 from the remaining 5 ie 5C2 =10

So totally there are 10 committees that include M.

Now coming to our favourable committees i.e. committees including M and A.
Since there are already two people out of the three needed the remaining person could be anyone from the remaining 4 people.

So there are 4 committees that include M and N.

So it is 4/10 = 40%
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Last edited by cicerone on 25 Sep 2008, 00:11, edited 1 time in total.
Senior Manager
Joined: 11 Jul 2006
Posts: 383
Location: TX
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Yes it should be C , here , the Key is the Question here
what percent of all the possible subcommittees that include Michael also include Anthony?
Then we have only 10 combination which include Michael in the committe.

6C3 includes the commitees which do not include Michael.
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