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

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Re: combination [#permalink] New post 17 Oct 2010, 18:11
E. I used the formula and got 8!/(2!2!2!2!)*4!
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Re: combination [#permalink] New post 18 Jan 2011, 09:52
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] New post 01 Jul 2011, 20:35
Nice question!....great explanations from Bunuel...
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Re: A group of 8 friends want to play doubles tennis. How many [#permalink] New post 14 Sep 2013, 12:27
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Re: A group of 8 friends want to play doubles tennis. How many [#permalink] New post 21 Sep 2013, 00:15
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*1


note: BOLD in even places .. just to distinguish among last three 1s
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Re: combination [#permalink] New post 19 Apr 2014, 17:50
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: combination-groups-and-that-stuff-85707.html#p642634

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}



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] New post 20 Apr 2014, 01:54
Expert's post
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: combination-groups-and-that-stuff-85707.html#p642634

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}



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: combination   [#permalink] 20 Apr 2014, 01:54
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