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

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Math Expert
Joined: 02 Sep 2009
Posts: 60480
Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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24 Feb 2012, 15:16
4
2
fortsill wrote:
Bunuel wrote:
Fourth approach:
Total # of splitting group of 6 into two groups of 3: 6C3*3C3/2!=10

Do you mind elaborating on that calculation, please? Isn't 6C3 the right calculation to select 3 people from a group of 6? You seem to be multiplying that by 3C3/2!

Is there another example that illustrates this - 6C3*3C3/2!- concept, thanks.

I guess your main concern is about dividing by 2! (because 3C3 is just selecting 3 out of 3 which is 1).

Dividing group of 6 objects {A, B, C, D, E, F} into 2 groups of 3: $$\frac{C^3_6*C^3_3}{2!}=10$$, we are dividing by $$2!$$ as there are 2 groups and order of these groups doesn't matter.

For example if we choose with $$C^3_6$$ the group {ABC} then the group {DEF} is left and we have two groups {ABC} and {DEF} but then we could choose also {DEF}, so in this case second group would be {ABC}, so we would have the same two groups: {ABC} and {DEF}. So to get rid of such duplications we should divide $$C^3_6*C^3_3$$ by factorial of number of groups - 2!.

Questions about the same concept to practice:
in-how-many-different-ways-can-a-group-of-8-people-be-125985.html
a-group-of-8-friends-want-to-play-doubles-tennis-how-many-106277.html
how-many-ways-are-there-to-split-a-group-of-6-boys-into-two-105381.html
a-group-of-8-friends-want-to-play-doubles-tennis-how-many-104807.html
in-how-many-different-ways-can-a-group-of-9-people-be-101722.html
in-how-many-different-ways-can-a-group-of-8-people-be-99053.html
nine-dogs-are-split-into-3-groups-to-pull-one-of-three-88685.html
in-how-many-different-ways-can-a-group-of-8-people-be-85707.html

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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24 Feb 2012, 15:31
@bunuel, how very generous with the explanations and the samples - terrific, thanks!
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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19 Jul 2012, 21:32
5
Totak ways to select 3 members from 6 members = 6C3 = 6!/3!*3! = 20
Now if the sis members are A,M,X1,X2,X3,X4 , take A & M together as a sing le unit Y.
Y = A,M, they can be selected as 2! ways = 2
Grups can be made as = YX1, YX2,YX3 & YX4 = 4
Total grups = 4*2=8

Now % is = 8/20*100=40% is the answer
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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, 14:45
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, 02:39
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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Posts: 2
Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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07 Dec 2015, 17:13
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, 04:32
3
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, 11:34
2
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, 16:25
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, 10:45
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!
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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, 11:00
1
jwang516 wrote:
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!

Note what the question is: what percent of all the possible subcommittees that include Michael also include Anthony

So out of all subcommittees that include Michael, how many include Anthony too?

All subcommittees that include Michael = 1*5C2 = 10
Out of these 10, how many will have Anthony too? Michael, Anthony and the third person can be picked in 4 ways (out of 4 remaining people) = 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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25 Nov 2019, 05:30
Bunuel Can you please explain the fourth approach in more detail? why have multiplied 6C3 with C3 and then divided it by 2!. Please explain the logic.
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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25 Nov 2019, 20:57
Bunuel 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 $$\frac{2}{5}=40\%$$.

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.

Bunuel
in your fourth approach, Can you please xplain why did we divide the combinatins by 2
Total # of splitting group of 6 into two groups of 3: $$\frac{C^3_6*C^_3}{2!}=10$$;
Re: Anthony and Michael sit on the six-member board of directors for compa   [#permalink] 25 Nov 2019, 20:57

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