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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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02 Nov 2005, 02:38
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Can someone explain how to tackle this problem? Why can't you simply use the combo formula? I am failing to see why?
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02 Nov 2005, 02:53
2C8x2C6x2C4x2C2=28x15x6x1=2520 . Select 2 out of 8, then 2 out of 6 ,then 2 out of 4 and finally two out of 2 . Multiply all and get 2520
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02 Nov 2005, 03:16
This question has been posted before.

Two ways to solve:

Combinations: C(8,2)C(6,2)C(4,2)C(2,2) / 4! = 105

Picking one player and evaluating the options from there: 7*5*3 = 105
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02 Nov 2005, 08:27
What is OA? I get 105 as well...7*5*3
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02 Nov 2005, 08:54
jdtomatito wrote:
This question has been posted before.

Combinations: C(8,2)C(6,2)C(4,2)C(2,2) / 4! = 105

I understood the second method. Could you please explain the first one as why we need to divide by 4!?
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02 Nov 2005, 09:14
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02 Nov 2005, 22:07
02 Nov 2005, 22:07
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