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A group of 8 friends want to play doubles tennis. How many
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10 Nov 2007, 20:43
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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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17 Feb 2010, 13:56
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, 22:35
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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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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Re: combination
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17 Feb 2010, 02:39
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, 10:52
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, 11:23
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, 11:45
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, 22:34
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, 08:55
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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Re: combination
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22 Feb 2010, 00:09
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, 09:09
honeyrai wrote: 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? The order of the groups matter in case we have team #1, #2, etc. Or in other words when we have specific assignment for each team. Consider the following: we have one set of three teams: A, B & C and three tasks: 1, 2, & 3. In how many ways can we assign these teams to these tasks? The answer would be 3!=6: 123ABC ACB BAC BCA CAB CBA Ian Stewart explains this very well at: combinationgroupsandthatstuff85707.html#p642634Hope it helps.
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Re: Hypo Combinations Question
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14 Nov 2010, 21:08
Hi how do we get the answer for the Ist question as 105. Pls explain



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A group of 8 friends want to play doubles tennis. How many
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15 Nov 2010, 00:57
AkritiMehta wrote: Hi how do we get the answer for the Ist question as 105. Pls explain 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 \(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105\), 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 You can check similar problems: http://gmatclub.com/forum/probability8 ... %20equallyhttp://gmatclub.com/forum/probability8 ... ide+groupshttp://gmatclub.com/forum/combination5 ... ml#p690842http://gmatclub.com/forum/subcommittee ... ide+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: How many different ways to play doubles tennis ?
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12 Jul 2011, 01:26
Alchemist1320 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 Formula = 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 important is (mn)!/n^m*m! so its 8!/(2^4*4!) = 105 = E



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Re: How many different ways to play doubles tennis ?
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12 Jul 2011, 02:42
Alchemist1320 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 From the question, we can say that the teams are not distinct i.e. we don't have team A, team B etc. But let's solve this question by making first team, second team, third team and fourth team. Later we will adjust the answer. Out of 8 people, how can you make the first team? In 8C2 ways. Out of 6 people, how can you make the second team? In 6C2 ways. Out of 4 people, how can you make the third team? In 4C2 ways. Out of 2 people, how can you make the fourth team? In 2C2 ways. How can you make the 4 teams? 8C2 *6C2*4C2*2C2 But here, we have considered the 4 teams to be distinct. We called them 'first team', 'second team' etc. So we need to divide the result by 4! to 'unarrange' them. You get: 8C2 *6C2*4C2*2C2/4! = 8*7*6*5*4*3*2*1/16*4! = 105
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Re: How many different ways to play doubles tennis ?
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14 Sep 2011, 13:20
karishma, Quote: But here, we have considered the 4 teams to be distinct. We called them 'first team', 'second team' etc. So we need to divide the result by 4! to 'unarrange' them.
Can you please explain this? I understand that we divide the slots! to remove identical stuff but here how does it make sense?



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Re: How many different ways to play doubles tennis ?
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14 Sep 2011, 21:07
shankar245 wrote: Can you please explain this? I understand that we divide the slots! to remove identical stuff but here how does it make sense? This is the logic behind this step: Say there are 4 boys: A, B, C, D There are two ways of splitting them in two groups. Method I The two groups can be made in the following ways 1. AB and CD 2. AC and BD 3. AD and BC The groups are not named/distinct. You have 4 boys in front of you and you split them in 2 groups and do not name the groups. There are 3 total ways of doing this. Method II On the other hand, I could put them in two distinct groups in the following ways 1. Group1: AB, Group2: CD 2. Group1: CD, Group2: AB (If you notice, this is the same as above, just that now AB is group 2) 3. Group1: AC, Group2: BD 4. Group1: BD, Group2: AC 5. Group1: AD, Group2: BC 6. Group1: BC, Group2: AD Here I have to put them in two different groups, group 1 and group 2. AB and CD is not just one way of splitting them. AB could be assigned to group 1 or group 2 so there are 2 cases. In this case, every 'way' we get above will have two possibilities so total number of ways will be twice. So there will be 6 total ways. Here since the groups are not distinct but 8C2 * 6C2 * 4C2 * 2C2 makes them distinct (we say, select the FIRST group in 8C2 ways, SECOND group in 6C2 ways etc), we need to divide by 4!.
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Re: combination
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20 Apr 2014, 01:54
Enael wrote: Bunuel wrote: jeeteshsingh 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 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 \(\frac{(mn)!}{(n!)^m}\) I tried using the formula and got: m = 4 groups n = 8 people (4*8)!/(8!)^4*4! but the result was way off. Am I using it wrongly? Appreciate the help. 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 \(\frac{(mn)!}{(n!)^m*m!}\).How many different ways can the group be divided into 4 teams (m) of 2 people (n)? \(\frac{(mn)!}{(n!)^m*m!}=\frac{8!}{(2!)^4*4!}=105\). Hope it helps.
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Re: A group of 8 friends want to play doubles tennis. How many
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27 May 2016, 22:20
Bunuel wrote: AkritiMehta wrote: Hi how do we get the answer for the Ist question as 105. Pls explain 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 \(\frac{C^2_8*C^2_6*C^2_2}{4!}=105\), 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 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. I do not agree that we need to divide it by 4! 8C2 * 6C2*4C2 is a regular combination formula We choose 2 persons out of 8. 1st team is formed. There is no arrangement, just selection. (8C2) We then choose 2 persons out of remaining 6. 2nd team is formed. There is no arrangement, just selection. (6C2) We then choose 2 persons out of remaining 4. 3rd team is formed. There is no arrangement, just selection. (4C2) We then choose 2 persons out of remaining 2. 4th team is formed. There is no arrangement, just selection. (2C2) Product: 8C2 * 6C2*4C2 This works out to 8! / (2!.2!.2!.2!) = 7!/2 = 2520




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