Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Want to score 90 percentile or higher on GMAT CR? Attend this free webinar to learn how to pre-think assumptions and solve the most challenging questions in less than 2 minutes.
A committee of three people is to be chosen from four teams
[#permalink]
Show Tags
12 Apr 2012, 18:48
3
25
00:00
A
B
C
D
E
Difficulty:
15% (low)
Question Stats:
78% (01:23) correct 22% (02:06) wrong based on 338 sessions
HideShow timer Statistics
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 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!!
Re: A committee of three people is to be chosen from four teams
[#permalink]
Show Tags
12 Apr 2012, 22:50
8
11
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 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]
Show Tags
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.
Re: A committee of three people is to be chosen from four teams
[#permalink]
Show Tags
02 Oct 2012, 08:49
1
Archit143 wrote:
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.
Re: A committee of three people is to be chosen from four teams
[#permalink]
Show Tags
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!
Re: A committee of three people is to be chosen from four teams
[#permalink]
Show Tags
03 Oct 2012, 03:07
vsprakash2003 wrote:
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: A committee of three people is to be chosen from four teams
[#permalink]
Show Tags
05 Oct 2012, 00:41
2
2
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!
_________________
If you find my post helpful, please GIVE ME SOME KUDOS!
Re: A committee of three people is to be chosen from four teams
[#permalink]
Show Tags
09 Mar 2016, 09:43
1
bapun11 wrote:
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!
Re: A committee of three people is to be chosen from four teams
[#permalink]
Show Tags
10 Mar 2016, 04:47
Here is how I solved this -
A/A1, B/B1, C/C1, D/D1
Total ways of choosing without any ristriction: 8C3 = 8!/3!5! = 56
Now total number of cases that are not required -
1. One couple is picked - 2C2 2. Remaining one member is picked from remaining 6 people - 6C1 3. Since there are 4 couple so this can be multiplied by 4.
Re: A committee of three people is to be chosen from four teams
[#permalink]
Show Tags
23 Sep 2019, 11:48
total pairs;; 8*6*4/3*2 ; 32 IMO E
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?
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!!
gmatclubot
Re: A committee of three people is to be chosen from four teams
[#permalink]
23 Sep 2019, 11:48
close
78% (01:23) correct 
