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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 05 Feb 2006, 21:35
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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?


20%
30%
40%
50%
60%
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 [#permalink] New post 05 Feb 2006, 21:56
Is it 40%?

Total possibilities = 6C3 = 20

A & M can be in one in committee in 4 ways. (A,M,X where X can be any of the other 4 members)

There are 2 such committees. Hence total num of committes in which A &M are together = 4*2 = 8

Ans = 8/20 * 100 = 40%
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 [#permalink] New post 05 Feb 2006, 22:00
:stab , i dont know how i thot 20% , but giddi is my hero and i am learning from him.. so i take wht u say sir!
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 [#permalink] New post 05 Feb 2006, 23:38
# of 3 people teams = 6C3 = 20

# of 3 people teams with Michael and Anthony = M+A+any remaining 4 -> 4 ways

% = 4/20 * 100% = 20%
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 [#permalink] New post 05 Feb 2006, 23:40
giddi77 wrote:
Is it 40%?

Total possibilities = 6C3 = 20

A & M can be in one in committee in 4 ways. (A,M,X where X can be any of the other 4 members)

There are 2 such committees. Hence total num of committes in which A &M are together = 4*2 = 8

Ans = 8/20 * 100 = 40%


I think you do not need to multiply by 2 as we're concerned with who's in the team, and not what position they hold in the team.
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 [#permalink] New post 06 Feb 2006, 05:27
all possible SUB COMS THAT INCLUDE M are 5C2=10 or 10 options.
When M and A are together then we have 4C1 or 4 options .
Then 4/10 of all subcommittees that include M also include A
So i get 40%
may be i am wrong
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 [#permalink] New post 06 Feb 2006, 06:44
number of ways that the two persons are included= 4C1=4

total number of ways if choose randomly = 6C3 =20

the percentage = 4/20 = 20%

Giddi, please verify the final answer with us...

Last edited by celiaXDN on 06 Feb 2006, 06:51, edited 2 times in total.
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 [#permalink] New post 06 Feb 2006, 07:06
Ywilfried the last time you did this question your answer was 40% and mine was 20%, remember it? You gave a nice explanation:

http://www.gmatclub.com/phpbb/viewtopic.php?t=24644

"# of possible to form 3 person team from 6 people = 6C3 = 20. That leaves another 20 combinations for the other team. Total = 40 possibilites (20 team A combination, 20 team B combination).

If Micheal is in team A, then Anthony has to be in team A, and team A will have only 4 ways to form (due to 4 people remaining).

But Micheal and Anthony can be in team A or team B. So % = (4/20)*2*100 = 40%"

So, we must multiply with 2!

Last edited by allabout on 06 Feb 2006, 07:08, edited 1 time in total.
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Should be 20 [#permalink] New post 06 Feb 2006, 07:07
Total possible combination = 6C3 = 20

Now let the members be A B C D E M
total number of 3-member-committe in which A and M are present= 4 [NOT 8]
because
(AMB , AMC, AMD, AME) is same as (MAB, MAC, MAD, MAE)

(4/20)*100 = 20 answer
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 [#permalink] New post 06 Feb 2006, 07:55
BG wrote:
all possible SUB COMS THAT INCLUDE M are 5C2=10 or 10 options.
When M and A are together then we have 4C1 or 4 options .
Then 4/10 of all subcommittees that include M also include A
So i get 40%
may be i am wrong


BG,

That is nice and quick way explanation..

OA is 40%
  [#permalink] 06 Feb 2006, 07:55
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