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Re: combination [#permalink]
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17 Oct 2010, 19:11
E. I used the formula and got 8!/(2!2!2!2!)*4!
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Re: combination [#permalink]
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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 [#permalink]
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01 Jul 2011, 21:35
Nice question!....great explanations from Bunuel...



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Re: A group of 8 friends want to play doubles tennis. How many [#permalink]
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14 Sep 2013, 13:27
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Re: A group of 8 friends want to play doubles tennis. How many [#permalink]
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21 Sep 2013, 01:15
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 well, this one was interesting but easy .. first person = anyone = 1 second = out of 7 remaining = 7 third person  anyone = 1 fourth = out of 5 remaining = 5 ...... so on answer = 1* 7*1* 5*1* 3*1* 1note: BOLD in even places .. just to distinguish among last three 1s
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Re: combination [#permalink]
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19 Apr 2014, 18:50
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 \(\frac{(mn)!}{(n!)^m}\) 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.



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Re: combination [#permalink]
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20 Apr 2014, 02:54
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 [#permalink]
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26 Apr 2015, 20:22
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Re: A group of 8 friends want to play doubles tennis. How many [#permalink]
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Re: A group of 8 friends want to play doubles tennis. How many [#permalink]
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16 Jun 2016, 23:44
For the first 2 people can be choosen out of 8 people,for the second team 2 people out of 6 people,for the third team 2 people out of 4 people and for the last team 2 people out of 2 and since the order of the team doesn’t matter,so we will divide it by 4!( if the teams are T1,T2,T3 and T4,then arrangement as T1 T2 T3 T4 or T4 T2 T3 T1 is the same) 8c2*6c2*4c2*2c2/4!=105



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Re: A group of 8 friends want to play doubles tennis. How many [#permalink]
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21 Jun 2016, 18:17
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 To choose 2 out of 8 people we have 8C2 There are 4 teams hence 4*(8C2) = 112 so went with nearest answer E. But what's wrong with above logic?
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Re: A group of 8 friends want to play doubles tennis. How many [#permalink]
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19 Aug 2016, 14:45
In case that the teams were DISTINCT i.e. {A,B}, {C,D}, {E,F}, {G,H} is NOT the same set as: {E,F}, {C,D}, {A,B}, {G,H} how the question would be written ?



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A group of 8 friends want to play doubles tennis. How many [#permalink]
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16 Oct 2016, 09:22
\(\frac{(8*7)}{2} * \frac{(6*5)}{2} * \frac{(4*3)}{2} * \frac{(2*1)}{2}/\)
you divide this whole expression by 4*3*2*1
You calculate, and find 105




A group of 8 friends want to play doubles tennis. How many
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