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

Question

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%

Show Answer

Official Answer

C

(A) 20%
(B) 30%
(C) 40%
(D) 50%
(E) 60%

Bunuel · Accepted Answer

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Bunuel · Answer

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%

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\%

Answer: C.

Hope it helps.

maratikus · Answer

The answer is C. Let's look at a committee where Mike belongs (he's person # 1) on that committee. When we choose person #2, there is a 1/5 probability that it's going to be Anthony (then it doesn't matter who the third person is), and 4/5 probability that it's going to be someone else (then person #3 is going to be Anthony with probability 1/4). Total probability = 1/5+4/5*1/4 = 2/5

srini123 · Answer

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%

Bunuel · Answer

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.

srini123 · Answer

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

Bunuel · Answer

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.

Barkatis · Answer

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.

Bunuel · Answer

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.

Barkatis · Answer

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 ?

Bunuel · Answer

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}.

praveenvino · Answer

Bunuel, can you advise why do we have to divide 6C3 * 3C3 by 2!, in the fourth approach ?

Bunuel · Answer

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

fortsill · Answer

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.

Bunuel · Answer

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.

bhavinshah5685 · Answer

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

GMATPill · Answer

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 the initial 5 remaining people, choose 2 to fill up the 2nd and 3rd slots)

=   / (5C2)

= 4 / 10

= 40%

Hope that helps.

mbaiseasy · Answer

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%

Answer: C

jegf1987 · Answer

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.

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

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