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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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This post was BOOKMARKED

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Difficulty:

95% (hard)

Question Stats:

35% (02:29) correct
65% (01:20) wrong based on 40 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?

Re: Anthony and Michael sit on the six-member board of directors [#permalink]
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% _________________

"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."

Re: Anthony and Michael sit on the six-member board of directors [#permalink]
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

"# 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.

Re: Anthony and Michael sit on the six-member board of directors [#permalink]
06 Feb 2006, 07:07

1

This post was BOOKMARKED

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)

Re: Anthony and Michael sit on the six-member board of directors [#permalink]
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

Re: Anthony and Michael sit on the six-member board of directors [#permalink]
08 Jun 2015, 05:48

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Re: Anthony and Michael sit on the six-member board of directors [#permalink]
08 Jun 2015, 05:49

mxr55820 wrote:

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

total possible combinations = 6C3 x 2 = 40 because 2 subcom have to be formed

Re: Anthony and Michael sit on the six-member board of directors [#permalink]
08 Jun 2015, 05:53

Expert's post

1

This post was BOOKMARKED

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%

First approach: Let's take the group with Michael: there is a place for two other members and one of them should be taken by Anthony, as there are total of 5 people left, hence there is probability of 2/5=40%.

Second approach: Again in Michael's group 2 places are left, # of selections of 2 out of 5 5C2=10 - total # of outcomes. Select Anthony - 1C1=1, select any third member out of 4 - 4C1=4, total # =1C1*4C1=4 - total # of winning outcomes. P=# of winning outcomes/# of outcomes=4/10=40%

Third approach: Michael's group: Select Anthony as a second member out of 5 - 1/5 and any other as a third one out of 4 left 4/4, total=1/5*4/4=1/5; Select any member but Anthony as second member out of 5 - 4/5 and Anthony as a third out of 4 left 1/4, total=4/5*1/4=1/5; Sum=1/5+1/5=2/5=40%

Fourth approach: Total # of splitting group of 6 into two groups of 3: 6C3*3C3/2!=10 # of groups with Michael and Anthony: 1C1*1C1*4C1=4 P=4/10=40%

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