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

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Joined: 20 Jun 2012
Posts: 90
Location: United States
Concentration: Finance, Operations
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Re: A group of 8 friends want to play doubles tennis. How many [#permalink]

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New post 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*1


note: BOLD in even places .. just to distinguish among last three 1s
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Re: combination [#permalink]

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New post 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: 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]

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New post 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: 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: A group of 8 friends want to play doubles tennis. How many [#permalink]

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New post 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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New post 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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New post 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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New post 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
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A group of 8 friends want to play doubles tennis. How many [#permalink]

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New post 10 Dec 2017, 11:06
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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


Let the 8 people be: A, B, C, D, E, F, G, and H

Take the task of creating the teams and break it into stages.

Stage 1: Select a partner for person A
There are 7 people to choose from, so we can complete stage 1 in 7 ways

ASIDE: There are now 6 people remaining. Each time we pair up two people (as we did in stage 1), we'll next focus on the remaining person who comes first ALPHABETICALLY.
For example, if we paired A with B in stage 1, the remaining people would be C, D, E, F, G and H. So, in the next stage, we'd find a partner for person C.
Likewise, if we paired A with E in stage 1, the remaining people would be B, C, D, F, G and H. So, in the next stage, we'd find a partner for person B.
And so on...

Stage 2: Select a partner for the remaining person who comes first ALPHABETICALLY
There are 5 people remaining, so we can complete this stage in 5 ways.

Stage 3: Select a partner for the remaining person who comes first ALPHABETICALLY
There are 3 people remaining, so we can complete this stage in 3 ways.

Stage 4: Select a partner for the remaining person who comes first ALPHABETICALLY
There is 1 person remaining, so we can complete this stage in 1 way.

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus create 4 pairings) in (7)(5)(3)(1) ways (= 105 ways)

Answer: E

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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