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

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

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14 Dec 2010, 16:32
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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

Please help with the explanation of the solution !!
[Reveal] Spoiler: OA

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Re: Combination and selection into team [#permalink]

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14 Dec 2010, 16:49
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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

Please help with the explanation of the solution !!

$$\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105$$, 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:
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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Re: Combination and selection into team [#permalink]

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14 Dec 2010, 16:56
Thanks Bunnel....

You are awsome...as usual ...
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Re: Combination and selection into team [#permalink]

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14 Dec 2010, 22:20
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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15 Nov 2013, 06:32
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Re: A group of 8 friends want to play doubles tennis. How many [#permalink]

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30 Nov 2014, 03:19
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Re: A group of 8 friends want to play doubles tennis. How many [#permalink]

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25 Mar 2016, 07:26
Hello from the GMAT Club BumpBot!

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Re: A group of 8 friends want to play doubles tennis. How many   [#permalink] 25 Mar 2016, 07:26
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