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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 16 Jul 2006, 00:26
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?

Need the fastest way!!!!
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 [#permalink] New post 16 Jul 2006, 02:04
For any subcommittee that Michael is on, two of the other five members are on the same commitee. Thus Anthony is on the same committee as Michael in 2/5 of the possible assignments..
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 [#permalink] New post 16 Jul 2006, 05:05
20 % ?

total number of ways of selecting 2 commities of 3 each from 6ppl = 6C3 = 20

Now michael and anthony are together the 3rd person in the committee is one of the 4 remaining so 4 ways

% 4/20 * 100 = 20
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 [#permalink] New post 16 Jul 2006, 06:21
prude_sb wrote:
20 % ?

total number of ways of selecting 2 commities of 3 each from 6ppl = 6C3 = 20

Now michael and anthony are together the 3rd person in the committee is one of the 4 remaining so 4 ways

% 4/20 * 100 = 20


But there are two committees, so 4*2/20=2/5
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 [#permalink] New post 16 Jul 2006, 12:04
kevincan wrote:
prude_sb wrote:
20 % ?

total number of ways of selecting 2 commities of 3 each from 6ppl = 6C3 = 20

Now michael and anthony are together the 3rd person in the committee is one of the 4 remaining so 4 ways

% 4/20 * 100 = 20


But there are two committees, so 4*2/20=2/5


are u saying 6c2 is the number of ways the first commitee can be set up and each time the remaining 3 form one comittee?

then the number of committes per say will be 6c2 x 2 but ways inwhich commites can be formed is 6c2.

now committees with both of them in could be 4 x2 . same group as a comitte for 2 functions

so answer = 2x4/6c2x2 = 20%


Can we assume that same group can be considered as 2 comittes? working on both the functions .


i am :confused
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 [#permalink] New post 16 Jul 2006, 19:43
kevincan wrote:
For any subcommittee that Michael is on, two of the other five members are on the same commitee. Thus Anthony is on the same committee as Michael in 2/5 of the possible assignments..


Thanks Kevin. I did not expect that short. You made it so simple.

OA is indeed 2/5 i.e 40%.
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 [#permalink] New post 16 Jul 2006, 21:04
Total # of such committee = 6C3 = 6!/3!3! = 20 ways
If Michael and Anthony are in the team, the last person can be any 4 of the remaining people.
P = 4/20 = 1/5 = 20%

But this is only for the firs team. We have to apply to team2, so total = 40%
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 [#permalink] New post 16 Jul 2006, 21:13
This is right. Good explanation.
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  [#permalink] 16 Jul 2006, 21:13
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