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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 sybcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?

total number of subcommitties that include Michael = No of ways to select Micahel (1C1) * No of ways to select other Member for First subcommitte (5C1) * No of ways to select 2 members for second subcommitte (4C2) * No of ways to select 3rd commites (2C2) =1C1*5C1*4C2*2C2

total number of subcommitties that include Michael and Anothony in one team. = No of ways to select Micahel (1C1) * No of ways to select other Member Anthony for First subcommitte (1C1) * No of ways to select 2 members for second subcommitte (4C2) * No of ways to select 3rd commites (2C2) =1C1*1C1*4C2*2C2

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 sybcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?

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

what if the question was like this "how many possible subcommittees can be made with both Michael and Anthony part of each one"

the original question says already Michael is part of a committee and we want Anthony to be part of it ..is it right..am i lost some where.

on the condition that michael is already part of committee we can apply conditional probability to make sure that anthony is also inn the committee.

X2suresh, your approach seems to be too complex and therefore contains a few mistakes:

1. When you choose 3d members you use 2C2, but you should use 2C1*1C1 because we have two possibilities rather than one.

2. In your use of complex approach you incorporate order. For example, when we choose 3 members of 6 the correct formula is 6C3 that does not equal 6C1*5C1*4C1, because in the last case you should exclude order with division by 3P3.

So, the correct approach is following:

1C1*5C2=10 - the number of committee members with Michael 1C1*1C1*4C1=4 - the number of committee members with Michael and Anthony _________________

X2suresh, your approach seems to be too complex and therefore contains a few mistakes:

1. When you choose 3d members you use 2C2, but you should use 2C1*1C1 because we have two possibilities rather than one.

2. In your use of complex approach you incorporate order. For example, when we choose 3 members of 6 the correct formula is 6C3 that does not equal 6C1*5C1*4C1, because in the last case you should exclude order with division by 3P3.

So, the correct approach is following:

1C1*5C2=10 - the number of committee members with Michael 1C1*1C1*4C1=4 - the number of committee members with Michael and Anthony

oops!!! sorry .. I misread the question. I thought three -- 2 person subcommities . istead of two - 3 person subcommities. I got it now. = (1C1* 1C1*4C1 * 3C3) / (1C1*5C2*3C3) = 4/10=40%

Thanks for your time.. _________________

Your attitude determines your altitude Smiling wins more friends than frowning

oops!!! sorry .. I misread the question. I thought three -- 2 person subcommities . istead of two - 3 person subcommities. I got it now. = (1C1* 1C1*4C1 * 3C3) / (1C1*5C2*3C3) = 4/10=40%

Thanks for your time..

You are welcome! Now I understand why you went so deep

Walker.. in the denominator you should be taking 6C3 / 2! ... I learnt this from one of your recent posts. 6C3 alone takes the order of the two committees into account. divide by 2! to eliminate that effect

This was wrong solution and I've corrected it after maratikus undeceived me. _________________

This was wrong solution and I've corrected it after maratikus undeceived me.

It would have been right if you had divided 6C3 by 2!. If A and M are together, then the that committee can be formed in only 4 ways (any one of the remaining 4).

Total number of ways of forming 2 committees of three persons from given 6 persons = 6C3/ 2!

I don't know if you remember the question asked in this very forum... "How many ways can you form three teams of 4 players from a group of 12 players". Your solution was 12C4*8C4*4C4/3! . You explained that 3! was to eliminate the repeat counting introduced in the numerator. Same principle here Wanted to drop a thanks. Your explanation taught me to tackle such questions.

I still have doubts in using 6C3/2P2 instead of 5C2. Let's consider example, where there are 9 persons (including Anthony and Michael) and 3 committees.

Right solutions: 7C1/8C2=7*2/8*7=1/4

Your reasoning: 7C1/(9C3/3P3)=7*3!*3!/9*8*7=1/2 _________________

I still have doubts in using 6C3/2P2 instead of 5C2. Let's consider example, where there are 9 persons (including Anthony and Michael) and 3 committees.

Right solutions: 7C1/8C2=7*2/8*7=1/4

Your reasoning: 7C1/(9C3/3P3)=7*3!*3!/9*8*7=1/2

No my reasoning is not 9C3/3P3. My reasoning is (9C3*6C3*3C3) / 3P3

If we have 9 persons and we need to form 3 committees with 3 persons in each committee, the first committee can be formed in 9C3 ways. For the second committee you have only 6 persons. So you can form the second committee in 6C3 ways. For the 3rd committee, you have only 3C3=1 way. So total 9C3*6C3*3C3. But this includes ordering of committees (not order of persons within a committee). So divide by 3!.

(9C3*6C3*3C3) / 3P3 answers to the question: In how many different ways can we divide 9 persons by 3 committees? But not how many different committees we can form with Michael. We can form 8C2 committees with Michael. _________________

(9C3*6C3*3C3) / 3P3 answers to the question: In how many different ways can we divide 9 persons by 3 committees? But not how many different committees we can form with Michael. We can form 8C2 committees with Michael.

Ah.. thats such a subtle difference.... so here u r trying to find the ratio (M&A together):(all with M) and not (M&A together):(all possible).

Yeah! You did force me think a bit as your answer, 40%, was correct, but I felt that something was wrong. Anyway, It is a good opportunity to recall GMAT quant _________________

Yeah! You did force me think a bit as your answer, 40%, was correct, but I felt that something was wrong. Anyway, It is a good opportunity to recall GMAT quant

Actually i completely missed the part of question that says "... subcommittees that include Michael also include Anthony" All this while i have been trying to solve for "all possible".

"read the question fully before jumping to solve" ...thats the lesson for me.

Total no of ways committee of 3 can be formed = 6C3 = 20 ways

Hence we need to choose 2 member (M+A as one + any one from W,X,Y,Z) committee from 5 members (considering M+A as single unit) available Hence 5C2 ways = 10 ways

10/20 = 50%.

Question asks for "what percent of all the possible subcommittees that include Michael also include Anthony?". -> M+A/M

You calculated the percent of all possible subcommittees that include both Michael and Anthony. -> M+A/all

The total no of ways committee of 3 can be formed is not relevant.

Possible subcommittees that include M: If M in 1st subcommittee-> M _ _ & _ _ _ If M in 2nd subcommittee-> _ _ _ & M _ _ (order within subcommittees is not relevant) -> 5C2 x 2

Possible subcommittees that include M that also include A: If M in 1st subcommittee-> M A _ & _ _ _ If M in 2nd subcommittee-> _ _ _ & M A _ -> 4C1 x 2

4C1 x 2 / 5C2 x 2 = 4C1/5C2 = 4/10 = 40% _________________

I'm giving the explanation..please tell if you guys agree...

The no. of ways to select 3 members out of 6 is 6C3. Now, we assume that M and A are on same commitee which this question wants (A and M to be on the same commitee).

Now, we're left with one position in that commitee and 4 guys r der..so 4C1 is the no. of ways that 1 guy can be selected.

So, ways poosible with M and A on the same commitee is 4C1/6C3. But since there are 2 commitees to be formed, the same logic can be applied to other committee as well..

So the answer is 2*4C1/6C3 = 2*4/20 = 40%..

Please let me know if I'm wrong at any point so that my doubts also get cleared.

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

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24 Feb 2012, 14:36

I was leaning towards 20%, but now, I seem to be leaning towards 40%.

My reasoning for 20%: Assume 6 people: M,A,1,2,3,4

Desired Outcomes: 4 (MA1, MA2, MA3, MA4) Total Outcomes: 6C3 = 20 Therefore, probability = 4/20 = 20%

However, the reason I'm now leaning towards 40% is this: Total Outcomes: 6C3 = 20 (same as above) Desired Outcomes: MA1, MA2, MA3, MA4 in subcommittee1 (4 outcomes) + MA1, MA2, MA3, MA4 in subcommittee2 (another 4 outcomes) = total 8 outcomes. Therefore, probability = 8/20 = 40%

Would appreciate some guidance by the moderators, thanks.

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

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24 Feb 2012, 14:40

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fortsill wrote:

I was leaning towards 20%, but now, I seem to be leaning towards 40%.

My reasoning for 20%: Assume 6 people: M,A,1,2,3,4

Desired Outcomes: 4 (MA1, MA2, MA3, MA4) Total Outcomes: 6C3 = 20 Therefore, probability = 4/20 = 20%

However, the reason I'm now leaning towards 40% is this: Total Outcomes: 6C3 = 20 (same as above) Desired Outcomes: MA1, MA2, MA3, MA4 in subcommittee1 (4 outcomes) + MA1, MA2, MA3, MA4 in subcommittee2 (another 4 outcomes) = total 8 outcomes. Therefore, probability = 8/20 = 40%

Would appreciate some guidance by the moderators, thanks.

Correct answer is indeed C (40%). I added the OA to the initial post.

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%

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

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24 Feb 2012, 14:58

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.

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

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24 Feb 2012, 15:16

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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!.

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