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A group of 8 friends want to play doubles tennis. How many
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10 Nov 2007, 21:43
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A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people? A. 420 B. 2520 C. 168 D. 90 E. 105
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Re: combination
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17 Feb 2010, 14:56
jeeteshsingh wrote: Balvinder wrote: A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?
A. 420 B. 2520 C. 168 D. 90 E. 105 8c2 x 6c2 x 4c2 = 2520 = B We should divide this by 4! > 2520/4!= 105, as the order of the teams does not matter. 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 You can heck this: combinationgroupsandthatstuff85707.html#p642634There 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 \(\frac{(mn)!}{(n!)^m}\)
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Re: combination
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10 Nov 2007, 23:35
Balvinder wrote: A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?
A. 420 B. 2520 C. 168 D. 90 E. 105
There's another topic asking this very question down below. Here's my answer:
E) 105.
Start with the question, how many ways can you divide a group of 4 into 2 teams. The answer is 3. Suppose our group is A B C D. Then the possible teams are:
AB CD
AC BD
AD BC
(It's 4C2/2, because determining one of the teams automatically determines the other one.)
Now consider the case of a group of 6 divided into 3 teams. Our group: A B C D E F. A has to be on some team, and there are 5 possibilities: AB, AC, AD, AE, or AF. For each of these possibilities, the rest of the members form a group of 4 divided into 2 teams. So the overall result is
5*(4C2/2) = 15.
Now we're up to our case. Group = A B C D E F G H. Again, A must be on a teamthere are 7 possibilities. For each of those possibilities, there are 15 ways of dividing the remaining 6 members into 3 teams. So total possibilities = 7 * 15 = 105.




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Re: combination
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10 Nov 2007, 23:11
Balvinder wrote: A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?
A. 420 B. 2520 C. 168 D. 90 E. 105
I was expecting an answer choice of 28. This should be a simple question asking for a value of 8c2, which is 28. Am I missing something?



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The first team can be selected in 8c2 ways, the second in 6c2, the third in 4c2 and the fourth in 2c2 ways.
Multiplying the no. of ways, we have
(8!)/16 = 2520 ways.
The answer is B.



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Walker, can you please explain why you are dividing 8C2 * 6C2 * 4C2 by 4!
i seem to understand 7*5*3 approach, but want to know your reasoning. thanks!



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CaspAreaGuy wrote: Walker, can you please explain why you are dividing 8C2 * 6C2 * 4C2 by 4! i seem to understand 7*5*3 approach, but want to know your reasoning. thanks!
4!=4P4  we should exclude arrangement of 4 pairs in different order.



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Re: combination
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28 Sep 2009, 10:48
A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?
A. 420 B. 2520 C. 168 D. 90 E. 105
Ans : = 8C2 * 6C2 * 4C2 = 2520
Ans is B



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Re: combination
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17 Feb 2010, 03:39
Balvinder wrote: A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?
A. 420 B. 2520 C. 168 D. 90 E. 105 8c2 x 6c2 x 4c2 = 2520 = B
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Re: combination
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20 Feb 2010, 11:52
Bunuel wrote: jeeteshsingh wrote: Balvinder wrote: A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?
A. 420 B. 2520 C. 168 D. 90 E. 105 8c2 x 6c2 x 4c2 = 2520 = B We should divide this by 4! > 2520/4!= 105, as the order of the teams does not matter. 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 You can heck this: combinationgroupsandthatstuff85707.html#p642634There 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 \(\frac{(mn)!}{(n!)^m}\) Bunuel... I guess I have started to feel lost when it comes to Combinations and Permutations. I thought since I have used Combination, I don't consider the order.. but as we per you even 8c2 x 6c2 x 4c2 has considered the order in it... THis confuses me... Could you please help me with this? For me I thought.. if I had used 8p2 x 6p2 x 4p2.... I would have considered the order too.... is that not the case?
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Re: combination
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20 Feb 2010, 12:23
jeeteshsingh wrote: Bunuel... I guess I have started to feel lost when it comes to Combinations and Permutations. I thought since I have used Combination, I don't consider the order.. but as we per you even 8c2 x 6c2 x 4c2 has considered the order in it... THis confuses me... Could you please help me with this? For me I thought.. if I had used 8p2 x 6p2 x 4p2.... I would have considered the order too.... is that not the case? It's not about the order of the players IN the group, with C you you excluded the duplications of these kind: {AB} and {BA} are excluded by C. It's about the order of the groups themselves. Set of teams: {AB}{CD}{EF}{GH} is the same set as: {E,F}, {C,D}, {A,B}, {G,H} and by your formula we'll get these two identical sets of teams. Basically one particular set of teams, let's say again: {AB}{CD}{EF}{GH}, will be chosen by your formula 24 times (4!=24) and we "need" only one from these 24, hence we should divide 8c2 x 6c2 x 4c2 by 4! to get rid of these duplications. Hope it's clear.
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Re: combination
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20 Feb 2010, 12:45
Bunuel wrote: jeeteshsingh wrote: Bunuel... I guess I have started to feel lost when it comes to Combinations and Permutations. I thought since I have used Combination, I don't consider the order.. but as we per you even 8c2 x 6c2 x 4c2 has considered the order in it... THis confuses me... Could you please help me with this? For me I thought.. if I had used 8p2 x 6p2 x 4p2.... I would have considered the order too.... is that not the case? It's not about the order of the players IN the group, with C you you excluded the duplications of these kind: {AB} and {BA} are excluded by C. It's about the order of the groups themselves. Set of teams: {AB}{CD}{EF}{GH} is the same set as: {E,F}, {C,D}, {A,B}, {G,H} and by your formula we'll get these two identical sets of teams. Basically one particular set of teams, let's say again: {AB}{CD}{EF}{GH}, will be chosen by your formula 24 times (4!=24) and we "need" only one from these 24, hence we should divide 8c2 x 6c2 x 4c2 by 4! to get rid of these duplications. Hope it's clear. Thanks a lot Bunuel!
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Re: combination
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20 Feb 2010, 23:34
Bunuel wrote: It's not about the order of the players IN the group, with C you you excluded the duplications of these kind: {AB} and {BA} are excluded by C.
It's about the order of the groups themselves. Set of teams: {AB}{CD}{EF}{GH} is the same set as: {E,F}, {C,D}, {A,B}, {G,H} and by your formula we'll get these two identical sets of teams.
Basically one particular set of teams, let's say again: {AB}{CD}{EF}{GH}, will be chosen by your formula 24 times (4!=24) and we "need" only one from these 24, hence we should divide 8c2 x 6c2 x 4c2 by 4! to get rid of these duplications.
Hope it's clear. You chose 4! because of 4 sets? right. If we had 5 teams we would have chosen 5! irrespective of the fact that from how many people its being chosen from?



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Re: combination
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21 Feb 2010, 09:55
honeyrai wrote: Bunuel wrote: It's not about the order of the players IN the group, with C you you excluded the duplications of these kind: {AB} and {BA} are excluded by C.
It's about the order of the groups themselves. Set of teams: {AB}{CD}{EF}{GH} is the same set as: {E,F}, {C,D}, {A,B}, {G,H} and by your formula we'll get these two identical sets of teams.
Basically one particular set of teams, let's say again: {AB}{CD}{EF}{GH}, will be chosen by your formula 24 times (4!=24) and we "need" only one from these 24, hence we should divide 8c2 x 6c2 x 4c2 by 4! to get rid of these duplications.
Hope it's clear. You chose 4! because of 4 sets? right. If we had 5 teams we would have chosen 5! irrespective of the fact that from how many people its being chosen from? Yes we are dividing by the factorial of the number of teams. You can also check following problems to practice: probability88685.html#p669025subcommittee86346.html#p647698probability85993.html#p644656combinationgroupsandthatstuff85707.html#p642634Hope 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: combination
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22 Feb 2010, 01:09
Bunuel wrote: honeyrai wrote: Bunuel wrote: It's not about the order of the players IN the group, with C you you excluded the duplications of these kind: {AB} and {BA} are excluded by C.
It's about the order of the groups themselves. Set of teams: {AB}{CD}{EF}{GH} is the same set as: {E,F}, {C,D}, {A,B}, {G,H} and by your formula we'll get these two identical sets of teams.
Basically one particular set of teams, let's say again: {AB}{CD}{EF}{GH}, will be chosen by your formula 24 times (4!=24) and we "need" only one from these 24, hence we should divide 8c2 x 6c2 x 4c2 by 4! to get rid of these duplications.
Hope it's clear. You chose 4! because of 4 sets? right. If we had 5 teams we would have chosen 5! irrespective of the fact that from how many people its being chosen from? Yes we are dividing by the factorial of the number of teams. You can also check following problems to practice: probability88685.html#p669025subcommittee86346.html#p647698probability85993.html#p644656combinationgroupsandthatstuff85707.html#p642634Hope it helps. Thanks for these superb posts! The next difficult part is how to judge whether the order is important or not? Can you shed some light? In the first post on your list of suggested posts, why is order important for 9 dogs to be divided into 3 groups of 3 members each?



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Re: combination
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22 Feb 2010, 10:09



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Re: combination
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05 Mar 2010, 00:54
Thanks for sharing your simple n elegant approach. Bunuel wrote: jeeteshsingh wrote: Balvinder wrote: A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?
A. 420 B. 2520 C. 168 D. 90 E. 105 8c2 x 6c2 x 4c2 = 2520 = B We should divide this by 4! > 2520/4!= 105, as the order of the teams does not matter. 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 You can heck this: combinationgroupsandthatstuff85707.html#p642634There 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 \(\frac{(mn)!}{(n!)^m}\)



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Re: combination
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15 Oct 2010, 10:18
Bunuel's explanations are great I found out what my mistake was



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Re: combination
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18 Jan 2011, 10:52
There is a proper formula to calculate this: (mn!/(n!)^m)*(1/m!) According to the question stem > mn=8; m=4; n=2 Hence > (8!/(2!)^4)*(1/4!)=105 The Ans. is E



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Re: combination
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01 Jul 2011, 21:35
Nice question!....great explanations from Bunuel...




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