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Anthony and Michael sit on the six-member board of directors [#permalink]
05 Feb 2006, 21:35

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Difficulty:

(N/A)

Question Stats:

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 & 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% _________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

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

"# 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.

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)

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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