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

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Manager
Joined: 31 May 2010
Posts: 70
A group of 8 friends want to play doubles tennis. How many  [#permalink]

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14 Dec 2010, 15:32
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Difficulty:

75% (hard)

Question Stats:

48% (01:42) correct 52% (01:56) wrong based on 243 sessions

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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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Joined: 02 Sep 2009
Posts: 53067
Re: Combination and selection into team  [#permalink]

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14 Dec 2010, 15:49
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saurabhgoel 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

$$\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105$$, 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, ...

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:
probability-88685.html?hilit=different%20items%20divided%20equally
probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
sub-committee-86346.html?highlight=divide+groups

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 $$\frac{(mn)!}{(n!)^m}$$

Hope it helps.
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Manager
Joined: 31 May 2010
Posts: 70
Re: Combination and selection into team  [#permalink]

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14 Dec 2010, 15:56
Thanks Bunnel....

You are awsome...as usual ...
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Director
Joined: 03 Sep 2006
Posts: 777
Re: Combination and selection into team  [#permalink]

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14 Dec 2010, 21:20
8C2*6C2*4C2*2C2

Above is same as mentioned by Bunuel
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.

the order of the teams does not matter. If we choose, say, these teams:

{A,B}, {C,D}, {E,F}, {G,H}

that's exactly the same set of teams as these:

{C,D}, {A,B}, {G,H}, {E,F}

Because the order of the teams themselves does not matter, we must divide by 4! = 24, the number of different orders we can put the four teams in, because all 24 different orders are in fact the same set of teams.
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Re: A group of 8 friends want to play doubles tennis. How many  [#permalink]

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13 Oct 2018, 16:59
saurabhgoel 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

The first team can be selected in 8C2 = (8 x 7)/2! = 28 ways.

The next team can be selected in 6C2 = (6 x 5)/2! = 15 ways.

The next team can be selected in 4C2 = (4 x 3)/2! = 6 ways.

The final team can be selected in 2C2 = 1 way.

However, since ORDER OF THE TEAMS DOES NOT MATTER, we need to divide the total number of ways to select the teams by 4! since we have 4 different teams. So we have:

(28 x 15 x 6)/4! = 2520/24 = 105

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Re: A group of 8 friends want to play doubles tennis. How many   [#permalink] 13 Oct 2018, 16:59
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