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Ps: combination

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Senior Manager
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Ps: combination [#permalink] New post 31 Oct 2005, 18:00
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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
VP
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 [#permalink] New post 31 Oct 2005, 18:34
B.

1st team can be made in 8C2 ways
2nd team pair is to chose from remaining 6 people: 6C2
similarly:
3rd team: 4C2
4th team: 2C2

total combinations: 8C2 * 6C2 * 4C2 * 2C2 = 2520
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 [#permalink] New post 31 Oct 2005, 23:14
Got the same answer as duttsit. 8C2*6C2*4C2*2C2=2520
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 [#permalink] New post 31 Oct 2005, 23:21
I am afraid, I need to relook at my explanation.
E.

There will be total: 2520 combinations for 4 teams.
However, we need to eliminate duplicate combinations.
For example, (1,2), (3,4), (5,6), (7,8) is same as (3,4),(1,2),(5,6),(7,8)
That is, order of teams not important. We, therefore, need to divide by 4! = 24

all distinct combinations = 2520 / 24 = 105
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 [#permalink] New post 01 Nov 2005, 09:29
Quote:
However, we need to eliminate duplicate combinations.
For example, (1,2), (3,4), (5,6), (7,8) is same as (3,4),(1,2),(5,6),(7,8)
That is, order of teams not important. We, therefore, need to divide by 4! = 24


Wow, great catch duttsit, I got 2520 too before realizing that I needed to divide by 24 to get 105.

105 it is.
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 [#permalink] New post 01 Nov 2005, 16:58
You are right duttsit, thanks for explaining it
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 [#permalink] New post 01 Nov 2005, 22:25
First Team: 8C2
Second Team: 6C2
Third Team: 4C2
First Team: 2C2

The answer is 8C2*6C2*4C2*2C2

Too lazy to calculate.
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Re: Ps: combination [#permalink] New post 01 Nov 2005, 22:34
gmacvik 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


Answer is E (8c2x6c2x4c2x2c2)/4! = 105

okay, going to bed.
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Re: Ps: combination   [#permalink] 01 Nov 2005, 22:34
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