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A committee of three people is to be chosen from four teams [#permalink]

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12 Apr 2012, 18:48

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A committee of three people is to be chosen from four teams of two. What is the number of different committees that can be chosen if no two people from the same team can be selected for the committee?

A. 20 B. 22 C. 26 D. 30 E. 32

The easy way to solve this problem is to say there are 4C3 ways to choose a team. (2C1)(2C1)(2C1)(2C0) to choose the players. Hence the solution is (4C3)(2C1)(2C1)(2C1)(2C0)=32.

You can also do this longer way: Since each member on the committee must be from a different team lets look at the converse: How many ways can people of the same team be on the committe and subtract them out. We can pick 3 people from 8, 8C3=56, this is the total number of combinations. We can find the number of ways the a two people from the same team are on the committee by

1) The paired team on the committee is the first team (2C2)(2C1)(2C0)(2C0)=2 (2C2)(2C0)(2C1)(2C0)=2 (2C2)(2C0)(2C0)(2C1)=2

2) The paired team on the committee is the second team

4) The paired team on the committie is the fourth team (2C1)(2C0)(2C0)(2C2)=2 (2C0)(2C1)(2C0)(2C2)=2 (2C0)(2C0)(2C1)(2C2)=2

All those two add up to 24. 56-24=32.

Here is my question, please help me answer below TWO APPROACHES COMBINATION========================== Now if I wanted to do find the number of combinations of pair of the same team on the committee using combination.

(4C1)=# of ways to pick one team where we will grab two people from (1C1)= # of ways to pick the two people from the same two people on the team (3C1)= # of ways to pick the last person from the other three teams (2C1)=# of ways to pick the person from the other group

Hence (4C1)(1C1)(3C1)(2C1)=24 . 56-24=32. Is my thinking correct here?

PERMUTATIONS================================

Now if I wanted to do find the number of combinations of pair of the same team on the committee using permutation. I have three slots for the three committee members. I have 8 people. I have 8 to choose from for the first slot. 1 person for the second slot, as it has it be the first slots team member. The last person can be any of the 6. Hence 8*1*6=48. But I did not consider the different arrangement's of the Pair Pair Nopair committee. I have 3!/2!=3 to jumble them. So I have 48*3=144 ways of picking two people from the same team on the committe. 56-144 is negative! What am I doing wrong here !! HELP Bunuel!! Help Anyone!! Any help would be appreciated!!

A committee of three people is to be chosen from four teams of two. What is the number of different committees that can be chosen if no two people from the same team can be selected for the committee?

A)20 B)22 C)26 D)30 E)32

I'm not sure what are you exactly doing in your last approach.

Anyway, if you want to solve with slot method the simpler solution would be: 8*6*4/3!=32, 8 ways to choose for the first slot, 6 ways to choose for the second slot, 4 ways to choose for the third slot and dividing by 3! to get rid of duplication.

Another way to solve this problem would be: \(C^3_4*2^3=32\), where \(C^3_4\) is # of ways to choose which 3 team members out of 4 will be represented in the committee and multiplying this by 2*2*2 since we can choose any of 2 members from each chosen team.

Re: A committee of three people is to be chosen from four teams [#permalink]

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14 Apr 2012, 11:11

Hey Bunuel,

What I am asking is if we wanted to use the permutations/ordered slot method to calculate the number of ways we can have a committie in which two of its members of the same team, How would we do so?

We have 8 members and a 3 slots. So total number of ways is 8*7*6, divided by 3! for double counts. 8*7*6/3!=56 .

Now what are the number of ways can we have two teams members on the committee? We can pick any of the 8 for the first slot. The second slot is reserved for the team member of the first slot, so 1 ways. And the last slot can be any of the remaining 6.

So 8*1*6. But we have to account for double counts. Do we divide by 3! or 2!? This is were I am stuck. If we divide by 2!, we get 8*1*6/2=24

56-24=32, our answer. But I don't understand why we divide by 2!??

I hope I was a little more clearer. Thank you again for responding Bunuel!

Re: Committee Combination Tough Problem [#permalink]

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02 Oct 2012, 04:36

Bunuel wrote:

alphabeta1234 wrote:

A committee of three people is to be chosen from four teams of two. What is the number of different committees that can be chosen if no two people from the same team can be selected for the committee?

A)20 B)22 C)26 D)30 E)32

I'm not sure what are you exactly dong in your last approach.

Anyway, if you want to solve with slot method the simpler solution would be: 8*6*4/3!=32, 8 ways to choose for the first slot, 6 ways to choose for the second slot, 4 ways to choose for the third slot and dividing by 3! to get rid of duplication.

Another way to solve this problem would be: \(C^3_4*2^3=32\), where \(C^3_4\) is # of ways to choose which 3 team members out of 4 will be represented in the committee and multiplying this by 2*2*2 since we can choose any of 2 members from each chosen team.

A committee of three people is to be chosen from four teams of two. What is the number of different committees that can be chosen if no two people from the same team can be selected for the committee?

A)20 B)22 C)26 D)30 E)32

I'm not sure what are you exactly dong in your last approach.

Anyway, if you want to solve with slot method the simpler solution would be: 8*6*4/3!=32, 8 ways to choose for the first slot, 6 ways to choose for the second slot, 4 ways to choose for the third slot and dividing by 3! to get rid of duplication.

Another way to solve this problem would be: \(C^3_4*2^3=32\), where \(C^3_4\) is # of ways to choose which 3 team members out of 4 will be represented in the committee and multiplying this by 2*2*2 since we can choose any of 2 members from each chosen team.

Re: A committee of three people is to be chosen from four teams [#permalink]

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02 Oct 2012, 12:31

Bunuel - What do you mean by selecting "3 teams" out of 4? The question states that 3 people form a committee (or team) and need to be selected from 4 teams of 2 people each. Archit is right. Shouldn't it be 8C3?

However the slotting method you mention is correct. 8*6*4/3!

Bunuel - What do you mean by selecting "3 teams" out of 4? The question states that 3 people form a committee (or team) and need to be selected from 4 teams of 2 people each. Archit is right. Shouldn't it be 8C3?

However the slotting method you mention is correct. 8*6*4/3!

There are 4 teams. Now, if we select 3 teams from those 4 and each will send one member then thee committee will have 3 members and no 2 members from the same team.

Hope it's clear.

P.S. Please follow the links in my post above for similar questions to practice.
_________________

Re: Committee Combination Tough Problem [#permalink]

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03 Oct 2012, 03:37

The question states no two person can be from the same team but if you multiply 2*2*2 considering each team giving 2 members...is it corect pls expalin

Re: A committee of three people is to be chosen from four teams [#permalink]

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03 Oct 2012, 03:47

A committee of three people is to be chosen from four teams of two. What is the number of different committees that can be chosen if no two people from the same team can be selected for the committee?

srry i knw the question is simple but wordings are making it complex or i am emphasizing more its clearly mentioned no two people from the same team is there any hidden meaning that you have assumed any 2 person from each of the 3 teams chosen

A committee of three people is to be chosen from four teams of two. What is the number of different committees that can be chosen if no two people from the same team can be selected for the committee?

srry i knw the question is simple but wordings are making it complex or i am emphasizing more its clearly mentioned no two people from the same team is there any hidden meaning that you have assumed any 2 person from each of the 3 teams chosen

Re: A committee of three people is to be chosen from four teams [#permalink]

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04 Oct 2012, 10:25

A committee of three people is to be chosen from four teams of two. What is the number of different committees that can be chosen if no two people from the same team can be selected for the committee?

The question states no two person can be from the same team but if you multiply 2*2*2 considering each team giving 2 members...is it corect pls explain. thanks

Re: A committee of three people is to be chosen from four teams [#permalink]

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05 Oct 2012, 00:41

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A committee of three people is to be chosen from four teams of two. What is the number of different committees that can be chosen if no two people from the same team can be selected for the committee?

A. 20 B. 22 C. 26 D. 30 E. 32

My Method

First we find out the amount of combinations of 3 person teams from a pool of 8. C(3/8) = 8!/5!*3! = 56 ways

Next I find the number of ways we CAN make a 3 person team using two from the same group, so

We take 2 people from a 2 person group C(2/2) and mutiply that by taking any 1 person from the remaining 6 C(1/6)

C(2/2)*C(1/6) = 6

This is the number of combinations by taking both parties from pair A. As we have a four pairs we must multiply this by 4 (4x6 = 24)

So there are 24 ways in which two people from the same pair can work together.

Finally we subtract this from the total number of combinations to find the number of groups where pairs do NOT work together....

56 - 24 = 32!
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Re: A committee of three people is to be chosen from four teams [#permalink]

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03 Jul 2014, 04:40

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Re: A committee of three people is to be chosen from four teams [#permalink]

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28 Jul 2015, 10:16

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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in the slot method why we divide by 3! to get rid of duplication? I am not sure about this concept. it would be great if any one explain

It is an arrangement question similar to finding arrangements of AABBB = 5!/ (2!*3!) where 2! and 3! are done to remove the duplication of A and B as all As and all Bs are the same. Had all As or all Bs be different (A1 A2) or (B1 B2 B3) , then the answer would have been = 5!

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