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In how many different ways can a group of 8 people be [#permalink]
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13 Aug 2010, 08:38
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In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? A. 90 B. 105 C. 168 D. 420 E. 2520
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Last edited by Bunuel on 24 Feb 2012, 15:05, edited 2 times in total.
Edited the question and added the OA



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Re: ways to divide?? [#permalink]
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13 Aug 2010, 08:59
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bibha wrote: In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? 90 105 168 420 2520 \(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105\), we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are  1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2... You can think about this in another way. For the first person we can pick a pair in 7 ways; For the second one in 5 ways (as two are already chosen); For the third one in 3 ways (as 4 people are already chosen); For the fourth one there is only one left. So we have 7*5*3*1=105 Answer: B. You can check similar problems: probability88685.html?hilit=different%20items%20divided%20equallyprobability85993.html?highlight=divide+groupscombination55369.html#p690842subcommittee86346.html?highlight=divide+groupsThere is also direct formula for this: 1. 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 \(\frac{(mn)!}{(n!)^m*m!}\). 2. 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 important is \(\frac{(mn)!}{(n!)^m}\) Hope it helps.
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Re: In how many different ways can a group of 8 people be [#permalink]
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24 Mar 2012, 12:13
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Another way to think about this:
How many ways can you arrange the following: T1 T1 T2 T2 T3 T3 T4 T4
That would be: 8!/(2!*2!*2!*2!)
Then also recall that we don't care about the differences between the teams, therefore
8!/(2!*2!*2!*2!*4!) = 105



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Re: ways to divide?? [#permalink]
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04 Sep 2012, 15:45
Bunuel wrote: bibha wrote: In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? 90 105 168 420 2520 \(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105\), we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are  1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2... You can think about this in another way. For the first person we can pick a pair in 7 ways; For the second one in 5 ways (as two are already chosen); For the third one in 3 ways (as 4 people are already chosen); For the fourth one there is only one left. So we have 7*5*3*1=105 Answer: B. There is also direct formula for this: 1. 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 \(\frac{(mn)!}{(n!)^m*m!}\). 2. 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 important is \(\frac{(mn)!}{(n!)^m}\) Hope it helps. I understand how to get the answer, but I'm wondering why we assume that the order of the teams doesn't matter. The question just asks how many ways you can group them. When should we assume something matters or doesn't matter when the question doesn't specify? And also, just to clarify, when we do 8!/2!2!2!2!4!, we are also not worrying about the order within each team either right i.e. (a,b)=(b,a)?



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Re: In how many different ways can a group of 8 people be [#permalink]
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04 Sep 2012, 22:35
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We don't worry about the order within a team or between teams, because there is nothing in the question that would make it necessary to do so.
The question needs to give us information, why we should differentiate. For example, if those teams are given numbers to or being seated on a bench.
Or if one of the team members is the captain.
But that's not the case here.



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Re: ways to divide?? [#permalink]
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05 Sep 2012, 00:39
dandarth1 wrote: Bunuel wrote: bibha wrote: In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? 90 105 168 420 2520 \(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105\), we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are  1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2... You can think about this in another way. For the first person we can pick a pair in 7 ways; For the second one in 5 ways (as two are already chosen); For the third one in 3 ways (as 4 people are already chosen); For the fourth one there is only one left. So we have 7*5*3*1=105 Answer: B. There is also direct formula for this: 1. 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 \(\frac{(mn)!}{(n!)^m*m!}\). 2. 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 important is \(\frac{(mn)!}{(n!)^m}\) Hope it helps. I understand how to get the answer, but I'm wondering why we assume that the order of the teams doesn't matter. The question just asks how many ways you can group them. When should we assume something matters or doesn't matter when the question doesn't specify? And also, just to clarify, when we do 8!/2!2!2!2!4!, we are also not worrying about the order within each team either right i.e. (a,b)=(b,a)? The teams are not numbered/labeled (we don't have team #1, #2, ...), the teams are not assigned to something (for example to tournaments), ... So, the order of the teams doesn't matter. Please check the links in my previous post for similar problems. Hope it helps.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: In how many different ways can a group of 8 people be [#permalink]
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14 Nov 2015, 03:29
bibha wrote: In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? A. 90 B. 105 C. 168 D. 420 E. 2520 Ways of choosing 2 out od a group of n people = nC2 i.e. Ways of choosing first team of 2 out of 8 = 8C2 i.e. Ways of choosing Second team of 2 out of remaining 6 = 6C2 i.e. Ways of choosing Third team of 2 out of remaining 4 = 4C2 Remaining 2 will form Forth team Total Ways of Choosing Teams = 8C2 * 6C2 * 4C2 * 1 BUT Since the first team may come on second places and second may come on third etc. i.e.e arrangement among teams are included here which we NEED to exclude i.e. Total Ways of Choosing Teams = 8C2 * 6C2 * 4C2 * 1 / 4! = 28*15*6/24 = 105 Answer: option B
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Re: In how many different ways can a group of 8 people be [#permalink]
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15 Oct 2016, 21:29
bibha wrote: In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? A. 90 B. 105 C. 168 D. 420 E. 2520 The FIRST pair can be chosen in 8C2 ways The SECOND pair can be chosen in 6C2 ways as 6 members are left after choosing 1st pair The THIRD pair can be chosen in 4C2 ways as 4 members are left after choosing 1st two pair and so on... Therefore total ways of splitting 8 members in 6 pairs = (8C2 * 6C2 * 4C2 * 2C2) / 4!The entire expression is divided by 4! because the arrangements of pair have been accounted for, which need to be excluded. Arrangement for example Same pair which came at first place can be chosen at third place or forth places also while we are choosing different pairs at different placesRequired answer = (28 * 15 * 6 * 1)/24 = 105 Answer: option C
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