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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]

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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%

OPEN DISCUSSION OF THIS QUESTION IS HERE: anthony-and-michael-sit-on-the-six-member-board-of-directors-102027.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 15 Jul 2013, 14:11, edited 2 times in total.
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05 Nov 2009, 07:20
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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%

Hope it helps.
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05 Nov 2009, 14:04
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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
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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05 Nov 2009, 13:35
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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 $$\frac{(mn)!}{(n!)^m*m!}$$.
$$\frac{6!}{(3!)^2*2!}=10$$

OR 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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05 Nov 2009, 07:47
bunuel, it will be more helpful if u can explain the fourth approach in detail
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05 Nov 2009, 07:50
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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05 Nov 2009, 09:18
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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 $$\frac{(mn)!}{(n!)^m*m!}$$.
$$\frac{6!}{(3!)^2*2!}=10$$

OR another way:
In our case $$\frac{6C3*3C3}{2!}=10.$$
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05 Nov 2009, 14:10
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
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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03 Aug 2012, 09:41
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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28 Oct 2012, 10:00
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]

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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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26 Oct 2013, 01:43
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
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.

I see what you mean now. What confused me about the fourth approach is that we are dividing the # of groups with Anthony, Michael and someone else BY # of different split-ups of 2 subcommittees (i understand why this is 10). I don't see how that's an apples to apples comparison? However, I can see it now that you've listed it out. I think it doesnt matter because if we want Anthony and Michael to be together they can only be in 4 split ups anyway?
Re: Sub-committee   [#permalink] 26 Oct 2013, 01:43
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