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Anthony and Michael sit on the six-member board of directors [#permalink]
02 Jan 2008, 13:50

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Difficulty:

85% (hard)

Question Stats:

43% (02:17) correct
57% (01:28) wrong based on 152 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. 20% B. 30% C. 40% D. 50% E. 60%

Can someone explain how to do this problem? My approach is 1. figure out total possible subcommittees 6C3 2. Numer of committees with M/A on it. MA4 = 4. So, its 4/15. But that is not correct. Can someone explain why?

6C3 = 6!/3!3! = 20 sub committees Michael will be included in 1/2 or 10 of these If Michael and Anthony are in a 3 person committee together there is only one spot left and 4 people left to fill it. 4C1 = 4!/3! = 4 4/10 = 40%

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%

Can someone explain how to do this problem? My approach is 1. figure out total possible subcommittees 6C3 2. Numer of committees with M/A on it. MA4 = 4. So, its 4/15. But that is not correct. Can someone explain why? OA = 40%

Simple. Actually at first I made a mistake I thought the total possibilities was 3!*3!, but really its 6!/3!*3! --> 20 possibilities.

then just count the possibilities: we have AMXYZW --> AMX, AMY, AMZ, AMW but we have to multiply this by 2 b/c two possible situations.

So total number of subcommittees that include Micheal = 5C2

Where do we get this 5?

There are 20 sub committees that can be made. More accurately, there are 10 PAIRS of sub committees since we have 6 people split up into 2 groups of 3. Michael can only be in 1 out of every pair of 2 groups. Thus, he is in 1/2 of the sub committees.

There are 2C10 combinations which can be added to Michael, which means 10. Michael+Anthony+x(or y or w or z); so we have 4 possibilities out of 10...40%..is that reasoning valid?

There are 2C10 combinations which can be added to Michael, which means 10. Michael+Anthony+x(or y or w or z); so we have 4 possibilities out of 10...40%..is that reasoning valid?

Where are you getting 2C10?

6C3 = number of subcommittees 6C3/2 = number of subcommittees Michael is in.

So Michael is in 10 of these subcommittees, how many can Anthony be in as well?

Assuming Anthony and Michael are in a committee together that leaves us with 2 of the 3 spots filled and 4 out of 6 people left to fill that last place. 4C1 = possible committees with both Anthony and Michael in them

If you fix Michael in one of the two subcommitees, there are two free chairs in that subcommitee and thre free chairs in the other one. Therefore the probability will be 2/5=40%

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%

Can someone explain how to do this problem? My approach is 1. figure out total possible subcommittees 6C3 2. Numer of committees with M/A on it. MA4 = 4. So, its 4/15. But that is not correct. Can someone explain why? OA = 40%

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%

Soln: Let the commitees be I and II Assuming that Anthony and Michael go into commitee I, there is just one more place left to be filled in that commitee and it can be taken by any of the 4 remaining people. Thus 4 ways. Since the Anthony and Michael can also go into commitee II, we get 4 ways for that commitee also. So in total = 8 ways

Now total number of ways of choosing 3 from 6 people is = 6C3 = 6 * 5 * 4/3! = 20 ways

solution=# of possible outcomes/# of total possibilities

# of possible outcomes = 2 ( _ _ _ ) = 2 ( M A _ ) [M & A could be anywhere ] = 2 ( 1c1 1c1 4c1) [1c1 = only one way to chose one place, 4c1= third place in a sub-committe can be occupied by any four of the remaining six people] = 2 ( 1 * 1 * 4 ) = 8

I said there is 6 slots. Three will be filled, three not filled.

654321 YYYNNN

Which gives me 120/6 = 20 Combos. But since it said two groups I multiplied that by 2.

20 * 2 = 40 (this is where I went wrong)

Did the same mistake in the beginning, but then realized that only the one committee matters and actually both, denominator and numerator would have to be multiplied by 2, which gives us still 40%.

We need to find out: a) what is the # of comittees that only include M? b) what is the number of comittees that include M AND A?

To get the answer, divide the result in b) by result in a)

a) M is already in place (_M_ ___ ___ ), so there are 2 available slots and 5 remaining people to fill these slots. 5!/2!(5-2)! = 10 combinations including M

b) M and A are in place (_M_ _A_ ___), so there is one available slot and 4 remaining people to fill it. 4!/1!(4-1)! = 4 combinations including M and A

We need to find out: a) what is the # of comittees that only include M? b) what is the number of comittees that include M AND A?

To get the answer, divide the result in b) by result in a)

a) M is already in place (_M_ ___ ___ ), so there are 2 available slots and 5 remaining people to fill these slots. 5!/2!(5-2)! = 10 combinations including M

b) M and A are in place (_M_ _A_ ___), so there is one available slot and 4 remaining people to fill it. 4!/1!(4-1)! = 4 combinations including M and A

I would like to learn if my approach to problem is correct. Below is the how I attacked this question.

1-) I assumed Ant. and Mich. as one person, so there might be 5!=120 possible group options.

2-) The problem says 2 subgroups; A, M, 4 3, 2, 1 4!= 24 subgroups Ant and Mich together

3-) another 4! since A and M might be in second subgroup too 4, 3, 2 A, M, 1

4-) 48/120 = 40 %

It seems long solution but it took less than 1 minute to apply. However I am not sure if this approach is right or I found the correct answer by coincidence. I appreciate and thank your comments in advance...

Re: Anthony and Michael sit on the six-member board of directors [#permalink]
21 Nov 2011, 15:18

aliensoybean wrote:

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%

Can someone explain how to do this problem? My approach is 1. figure out total possible subcommittees 6C3 2. Numer of committees with M/A on it. MA4 = 4. So, its 4/15. But that is not correct. Can someone explain why? OA = 40%

I feel like you guys are making this more difficult than necessary. I read it like this:

Anthony was picked to be on a subcommittee. There are two spots left on his subcommittee. Michael and four other guys are left to fill those two spots. What's the chance that Michael will be on Anthony's subcommittee? Answer: Michael's got 2 chances out of 5 guys, or 2/5, or 40%.

gmatclubot

Re: Anthony and Michael sit on the six-member board of directors
[#permalink]
21 Nov 2011, 15:18

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