Jill is dividing her ten-person class into two teams of eq

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Jill is dividing her ten-person class into two teams of eq [#permalink]

28 Feb 2013, 05:25

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42% (02:17) correct

57% (01:12) wrong

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Jill is dividing her ten-person class into two teams of equal size for a basketball game. If no one will sit out, how many different match-ups between the two teams are possible?

A. 10
B. 25
C. 126
D. 252
E. 630

[Reveal] Spoiler: OA

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Last edited by Bunuel on 28 Feb 2013, 06:14, edited 1 time in total.

Edited the question.

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Re: Jill is dividing her ten-person class into two teams of eq [#permalink]

28 Feb 2013, 06:29

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emmak wrote:

There should be 5 people in each group. We can divide a group of 10 people into 2 teams of 5 in \frac{C^5_{10}*C^5_5}{2!}=126 ways (dividing by 2! because the order of the groups doesn't matter).

Answer: C.

For more on this check:
a-group-of-8-friends-want-to-play-doubles-tennis-how-many-55369.html
in-how-many-different-ways-can-a-group-of-9-people-be-85993.html
probability-88685.html
probability-85993.html
combination-55369.html
sub-committee-86346.html

Hope it helps.
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Re: Jill is dividing her ten-person class into two teams of eq [#permalink]

02 Mar 2013, 11:47

here is a formula -http://gmatclub.com/forum/a-group-of-8-friends-want-to-play-doubles-tennis-how-many-55369.html#p689312
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 -
(mn)!/(n!)^m*m!

10! /((5!)^2*2!) =126
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Re: Jill is dividing her ten-person class into two teams of eq [#permalink] 02 Mar 2013, 11:47

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Jill is dividing her ten-person class into two teams of eq

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Jill is dividing her ten-person class into two teams of eq

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