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

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Intern
Joined: 23 Jun 2005
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Anthony and Michael sit on the six-member board of directors [#permalink]  14 Aug 2006, 19:52
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%

I dont get the explanation at all. Can someone explain this better?
CEO
Joined: 20 Nov 2005
Posts: 2919
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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Total subcommittees = 6C3 = 20

Subcommittees that include michael = 5C2 = 10
Subcommittees that include both = 4C1 = 4

_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

CEO
Joined: 20 Nov 2005
Posts: 2919
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 17

Kudos [?]: 87 [0], given: 0

GK_Gmat wrote:
ps_dahiya wrote:
Total subcommittees = 6C3 = 20

Subcommittees that include michael = 5C2 = 10 --> How? Please explain
Subcommittees that include both = 4C1 = 4 --> How? Please explain

If Michael is selected then we need to select two people out of 5. So its 5C2

If Michael and Anthony are selected then we need to select one more person out of 4 people. So its 4C1

Hoep this helps.
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Director
Joined: 09 Aug 2006
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ps_dahiya wrote:
GK_Gmat wrote:
ps_dahiya wrote:
Total subcommittees = 6C3 = 20

Subcommittees that include michael = 5C2 = 10 --> How? Please explain
Subcommittees that include both = 4C1 = 4 --> How? Please explain

If Michael is selected then we need to select two people out of 5. So its 5C2

If Michael and Anthony are selected then we need to select one more person out of 4 people. So its 4C1

Hoep this helps.

Great. Thanks a lot.
Intern
Joined: 09 Aug 2006
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I think another way to solve it is like this:

The probability for both Michael and Anthony to be selected to the first subcomittee is:
2/6 * (1/5) * (3C2) = 1/5

The total probability for both of them to be selected to either the first or second subcomittee is:
1/5*2 = 2/5 = 40%
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