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Anthony and Michael sit on the six-member board of directors [#permalink]
 04 Nov 2009, 17:28
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Anthony and Michael sit on the six member board od directors for compnay 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%
Last edited by Bunuel on 04 Aug 2012, 01:27, edited 1 time in total.
OA added.
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Anthony and Michael sit on the six member board od directors for compnay 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% Answer: C. Hope it helps.
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Bunuel wrote: srini123 wrote: Bunuel - the fourth approach you posted seems to have an error:
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%
Total # of splitting group of 6 into two groups of 3: should be 6C3*3C3=20 and not 10.
Out of these 20 , there are 10 groups with Michael in it
so answer = groups with michael and anthony/ total groups with michael = 4/10 = 40% Do we have group #1 and #2? Does it matter in which group Michael and Anthony will be? If no we should divide 6C3*3C3 by 2! and get 10 different split ups of 6 to groups of 3 when order of groups doesn't matter. # of groups with Michael and Anthony together: 1C1*1C1*4C1=4 P=4/10=40%. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is \frac{(mn)!}{(n!)^m*m!}. \frac{6!}{(3!)^2*2!}=10OR another way: In our case \frac{6C3*3C3}{2!}=10.Sorry I may be missing something but, if we want to split 6 persons into 2 groups of 3 where order doesn't matter. dont we have 20 groups ? for eg., let A,B,C,D,E,F be 6 people and the groups with 3 people in each will be as below ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF - 10 BCD, BCE,BCE, BDE,BDF, BEF - 6 CDE, CDE,CEF -3 DEF -1 total 20. Let A be michael for this question , then total groups we have with michael is 10 and 4 groups have Michael and Anthony (assume D=anthony) im sorry im still not getting why 20 is not the total groups irrespective of michael in it or not
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srini123 wrote: Sorry I may be missing something but, if we want to split 6 persons into 2 groups of 3 where order doesn't matter. dont we have 20 groups ?
for eg., let A,B,C,D,E,F be 6 people and the
groups with 3 people in each will be as below
ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF - 10
BCD, BCE,BCE, BDE,BDF, BEF - 6
CDE, CDE,CEF -3
DEF -1
total 20.
Let A be michael for this question , then total groups we have with michael is 10 and 4 groups have Michael and Anthony (assume D=anthony)
im sorry im still not getting why 20 is not the total groups irrespective of michael in it or not Now I see what you've meant. But I think it's not quite true and here is where you are making the mistake: you listed # of selections of three out of 6 - 6C3=20. This is different from splitting the group into TWO groups of three members each: 1. ABC - DEF 2. ABD - CEF 3. ABE - CDF 4. ABF - CDE 5. ACD - BEF 6. ACE - BDF 7. ACF - BDE 8. ADE - BCF 9. ADF - BCE 10. AEF - BCD So, here are all possible ways to split 6 people into two groups of three members each. Let's say Anthony is A and Michael is B. You can see that there are only 4 scenarios when A and B are in one group (1,2,3,4). Hence 4/10.
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bunuel, it will be more helpful if u can explain the fourth approach in detail
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Bunuel - the fourth approach you posted seems to have an error: 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% Total # of splitting group of 6 into two groups of 3: should be 6C3*3C3=20 and not 10. Out of these 20 , there are 10 groups with Michael in it so answer = groups with michael and anthony/ total groups with michael = 4/10 = 40%
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srini123 wrote: Bunuel - the fourth approach you posted seems to have an error:
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%
Total # of splitting group of 6 into two groups of 3: should be 6C3*3C3=20 and not 10.
Out of these 20 , there are 10 groups with Michael in it
so answer = groups with michael and anthony/ total groups with michael = 4/10 = 40% Do we have group #1 and #2? Does it matter in which group Michael and Anthony will be? If no we should divide 6C3*3C3 by 2! and get 10 different split ups of 6 to groups of 3 when order of groups doesn't matter. # of groups with Michael and Anthony together: 1C1*1C1*4C1=4 P=4/10=40%. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is \frac{(mn)!}{(n!)^m*m!}. \frac{6!}{(3!)^2*2!}=10OR another way: In our case \frac{6C3*3C3}{2!}=10.
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Bunuel wrote: srini123 wrote: Sorry I may be missing something but, if we want to split 6 persons into 2 groups of 3 where order doesn't matter. dont we have 20 groups ?
for eg., let A,B,C,D,E,F be 6 people and the
groups with 3 people in each will be as below
ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF - 10
BCD, BCE,BCE, BDE,BDF, BEF - 6
CDE, CDE,CEF -3
DEF -1
total 20.
Let A be michael for this question , then total groups we have with michael is 10 and 4 groups have Michael and Anthony (assume D=anthony)
im sorry im still not getting why 20 is not the total groups irrespective of michael in it or not Now I see what you've meant. But I think it's not quite true and here is where you are making the mistake: you listed # of selections of three out of 6 - 6C3=20. This is different from splitting the group into TWO groups of three members each: 1. ABC - DEF 2. ABD - CEF 3. ABE - CDF 4. ABF - CDE 5. ACD - BEF 6. ACE - BDF 7. ACF - BDE 8. ADE - BCF 9. ADF - BCE 10. AEF - BCD So, here are all possible ways to split 6 people into two groups of three members each. Let's say Anthony is A and Michael is B. You can see that there are only 4 scenarios when A and B are in one group (1,2,3,4). Hence 4/10. Ok great, thanks Bunuel now I see splitting group of 6 into 2 groups with 3 in each is different from forming groups of 3 people into 2 groups from 6 people. Thanks much
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Bunuel wrote: arora2m 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?
1. 20%
2. 30%
3. 40%
4. 50%
5. 60% 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%. Bunuel, Can you please help with the above approach? I was able to understand the three other approaches outlined by you. Thanks
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voodoochild wrote: Bunuel wrote: arora2m 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?
1. 20%
2. 30%
3. 40%
4. 50%
5. 60% 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%. Bunuel, Can you please help with the above approach? I was able to understand the three other approaches outlined by you. Thanks I didn't get this explanation straight away too. I think the rationale is the following: Mike's probability to be in the sub-c A is 50% or 3/6, or 1/2. Anthony's probability to get into sub-c A is 2/5 or 40%. Probability that both of them are in sub-c A is (1/2)*(2/5)=1/5. Same applies to the sub-c B, hence 1/5*2= 2/5 or 40%.
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Re: Anthony and Michael sit on the six-member board of directors [#permalink]
30 Oct 2012, 14:45
Here's some clarification for those who are still confused. First group of 6| | | | | | Gets broken into 2 subcommittees of 3 each:| | | | | | We know Michael is already in one of them: M | | | | | We just need to fill in one of those two slots next to Michael with an "A" for Anthony.So we already know Mike is in that slot and that there are 5 remaining choices. Well, how many ways can we pick Anthony such that he ends up in Michael's group? Keep in mind that order matters - meaning Michael in the 1st slot is counted separately from Michael in the 2nd slot, etc. so we can multiply a line of nCr formulas: [ (Out of 1 available Michael, pick that 1 Anthony for that 2nd spot) * (Out of the remaining 4, choose any 1 for that 3rd spot) ]
= --------------------------------------------------------------------------------------------------------------------------------------------------------------------
(Out of the initial 5 remaining people, choose 2 to fill up the 2nd and 3rd slots) = [ (1C1) * (4C1) ] / (5C2) = 4 / 10 = 40%Hope that helps.
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Re: Anthony and Michael sit on the six-member board of directors
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30 Oct 2012, 14:45
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