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# Anthony and Michael sit on the six-member board of directors for compa

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

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Updated on: 12 Dec 2018, 00:01
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Question Stats:

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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%

Originally posted by arora2m on 04 Nov 2009, 16:28.
Last edited by Bunuel on 12 Dec 2018, 00:01, edited 3 times in total.
Renamed the topic and edited the question.
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Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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01 Oct 2010, 07:42
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Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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05 Nov 2009, 06:20
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9
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 $$\frac{2}{5}=40\%$$.

Second approach:

Again in Michael's group 2 places are left, # of selections of 2 out of 5 is $$C^2_5=10$$ = total # of outcomes.

Select Anthony - $$C^1_1=1$$, select any third member out of 4 - $$C^1_4=4$$, total # $$=C^1_1*C^1_4=4$$ - total # of winning outcomes.

$$P=\frac{# \ of \ winning \ outcomes}{total \ # \ of \ outcomes}=\frac{4}{10}=40\%$$

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 $$=\frac{1}{5}*\frac{4}{4}=\frac{1}{5}$$;

Select any member but Anthony as second member out of 5 - 4/5 and Anthony as a third out of 4 left 1/4, $$total=\frac{4}{5}*\frac{1}{4}=\frac{1}{5}$$;

$$Sum=\frac{1}{5}+\frac{1}{5}=\frac{2}{5}=40\%$$

Fourth approach:

Total # of splitting group of 6 into two groups of 3: $$\frac{C^3_6*C^_3}{2!}=10$$;

# of groups with Michael and Anthony: $$C^1_1*C^1_1*C^1_4=4$$.

$$P=\frac{4}{10}=40\%$$

Hope it helps.
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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05 Nov 2009, 06:50
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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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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05 Nov 2009, 08:18
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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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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05 Nov 2009, 12:35
1
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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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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05 Nov 2009, 13:04
2
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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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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01 Oct 2010, 08:05
Thanks.
Just a question: I noticed that you responded very quickly to my posts. Which I thank you for. But I am wondering if it's ok that I post questions that have been obviously posted before. If it's a problem then please tell me how can I check whether a question is already on the forum or not. Thanks.
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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01 Oct 2010, 08:26
2
Barkatis wrote:
Thanks.
Just a question: I noticed that you responded very quickly to my posts. Which I thank you for. But I am wondering if it's ok that I post questions that have been obviously posted before. If it's a problem then please tell me how can I check whether a question is already on the forum or not. Thanks.

Generally it's a good idea to do the search before posting, for example I would search in PS subforum (there is a search field above the topics) for the word "Anthony", don't think that there are many questions with this name. Though it's not a probelm to post a question that was posted before: if moderators find previous discussions they will merge the topics, copy the solution from there or give a link to it.
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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04 Oct 2010, 11:56
Bunuel, A question just came to my mind concerning the second approach of the solution that you proposed:

total # of winning outcomes
Select Anthony: 1C1=1, select any third member out of 4: 4C1=4, total # =1C1*4C1=4

If we say 1C1*4C1 don't we assume that the winning can be either (Anthony,Third member) or (Third member, Anthony) but not both ?
shouldn't we multiply 1C1*4C1 by 2 to get the total number of possible combinations ?
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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04 Oct 2010, 12:12
Barkatis wrote:
Bunuel, A question just came to my mind concerning the second approach of the solution that you proposed:

total # of winning outcomes
Select Anthony: 1C1=1, select any third member out of 4: 4C1=4, total # =1C1*4C1=4

If we say 1C1*4C1 don't we assume that the winning can be either (Anthony,Third member) or (Third member, Anthony) but not both ?
shouldn't we multiply 1C1*4C1 by 2 to get the total number of possible combinations ?

We are counting # of committees with Anthony and Michael:

{M,A,1};
{M,A,2};
{M,A,3};
{M,A,4}.

Here {M,A,1} is the same committee as {M,1,A}.
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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21 Jan 2011, 15:54
Bunuel, can you advise why do we have to divide 6C3 * 3C3 by 2!, in the fourth approach ?
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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21 Jan 2011, 16:14
1
6
praveenvino wrote:
Bunuel, can you advise why do we have to divide 6C3 * 3C3 by 2!, in the fourth approach ?

We are dividing by 2! (factorial of the # of groups) because the order of the groups is not important (we don't have the group #1 and the group #2) and we need to get rid of the duplications.

Dividing a group into subgroups:
combinations-problems-95344.html
split-the-group-101813.html
9-people-and-combinatorics-101722.html
ways-to-divide-99053.html
combination-and-selection-into-team-106277.html
ways-to-split-a-group-of-6-boys-into-two-groups-of-3-boys-ea-105381.html
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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30 Oct 2012, 13:45
1
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 for compa  [#permalink]

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28 Dec 2012, 01:39
3
Barkatis 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?

A. 20%
B. 30%
C. 40%
D. 50%
E. 60%

Here is my approach:

I first counted the possible creation of 2 subcommittees without restriction: $$\frac{6!}{3!3!}* \frac{3!}{3!}= 20$$
Then, I now proceed to counting the number of ways to create committee with Michael and Anthony together.

M A _ + _ _ _ = $$\frac{4!}{1!3!} * \frac{1!}{1!} = 4$$

MA could be in group#1 or group#2. Thus, $$=4*2 = 8$$

Final calculation: $$8/20 = 4/10 = 40%$$

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

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07 Dec 2015, 16:13
Bunuel wrote:
Barkatis 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?

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.

Hi Bunnuel,

On approach 3, are you implying that the order does matter? By calculating the odds of picking Anthony in the second position and anyone else in the third position AND the odds of picking anyone except Anthony in the second position and Anthony in the 3rd position you are basically establishing that the order does matter (whether Anthony is in the second or 3rd position).

Can you clarify? I am sure there is something wronog in my thought process. Thanks.
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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17 Dec 2015, 03:32
1
jegf1987 wrote:
Bunuel wrote:
Barkatis 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?

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.

Hi Bunnuel,

On approach 3, are you implying that the order does matter? By calculating the odds of picking Anthony in the second position and anyone else in the third position AND the odds of picking anyone except Anthony in the second position and Anthony in the 3rd position you are basically establishing that the order does matter (whether Anthony is in the second or 3rd position).

Can you clarify? I am sure there is something wronog in my thought process. Thanks.

Responding to a pm:

No, we know very well that order does not matter when forming groups. Here, you have to form a group/subcommittee so the order in which you pick people is irrelevant. That said, we consider order in method 3 because of the limitations of this particular method. We are using probability. I can find the probability that the "next" guy I pick is Anthony. But how do I find the probability that Anthony is one of the next two guys I pick? For that, I have to use two steps:
- The next one is Anthony or
- Next to next one is Anthony
We add these two probabilities and get the probability that either of the next two guys is Anthony.
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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15 Jun 2016, 10:34
Top Contributor
Barkatis 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?

A. 20%
B. 30%
C. 40%
D. 50%
E. 60%

We can think of this as a probability question: What is the probability that Anthony and Michael are on the same subcommittee?

Now, assume that we're creating subcommittees.
We want to place 6 people in the following spaces:
_ _ _ | _ _ _

First, we place Michael in one subcommittee; it makes no difference which one:
M _ _ | _ _ _

When we go to place Anthony, we see that there are 5 spaces remaining.
2 spaces are on the same subcommittee as Michael.
So the probability that they are on the same subcommittee is 2/5 = 40%

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

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14 Jun 2017, 15:25
Barkatis 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?

A. 20%
B. 30%
C. 40%
D. 50%
E. 60%

Let’s first determine the number of ways 2 three-person subcommittees can be formed from 6 people. The number of ways 3 people can be selected from 6 people for the first committee is 6C3 = (6 x 5 x 4)/(3 x 2) = 20. The number of ways 3 people can be selected from remaining 3 people for the second committee is 3C3 = 1. Thus, the number of ways 2 three-person subcommittees can be formed from 6 people is 20 x 1 = 20, if the order of selecting the committees matters. However, since the order of selecting the committees doesn’t matter, we have to divide by 2! = 2. Thus, the number of ways 2 three-person subcommittees can be formed from 6 people is 20/2 = 10.

Since only a total 10 committees can be formed, we can list all of these committees and see how many of them have Anthony and Michael on the same committee. We can let A be Anthony, M be Michae,l and B, C, D, and E be the other 4 people.

1) A-B-C, D-E-M
2) A-B-D, C-E-M
3) A-B-E, C-D-M
4) A-B-M, C-D-E
5) A-C-D, B-E-M
6) A-C-E, B-D-M
7) A-C-M, B-D-E
8) A-D-E, B-C-M
9) A-D-M, B-C-E
10) A-E-M, B-C-D

We can see that from the 10 committees that can be formed, 4 of them (in bold) include both Anthony and Michael. Thus, the probability is 4/10 = 40%

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

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15 Jun 2017, 09:45
I understand why 40% is the right answer but I have a question on why this approach doesn't work

Considering there are 2 subcommities A and B
Both Anthony and Michael can be in subcommity A or B

Therefore, we have:
Anthony in A and Michael in A
Anthony in B and Michael in A
Anthony in A and Michael in B
Anthony in B and Michael in B

They are in the same group in 2 of the 4 situations. Therefore 50%.

What part of this logic here is incorrect? Thanks!
Re: Anthony and Michael sit on the six-member board of directors for compa &nbs [#permalink] 15 Jun 2017, 09:45

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