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

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Joined: 09 Feb 2013
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Jill is dividing her ten-person class into two teams of eq [#permalink]

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28 Feb 2013, 05:25
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Difficulty:

95% (hard)

Question Stats:

40% (02:25) correct 60% (01:27) wrong based on 126 sessions

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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]

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28 Feb 2013, 06:29
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emmak wrote:
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

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).

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]

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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]

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22 Apr 2015, 03:02
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Re: Jill is dividing her ten-person class into two teams of eq [#permalink]

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23 Apr 2015, 15:42
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Hi All,

This question tests your knowledge of the Combination Formula, but it comes with a rare "twist" that most people don't realize. Here's a bit more information on that "twist":

With 10 players, the process of figuring out how many groups of 5 can be formed is pretty straight-forward....

10C5 = 10!/(5!5!) = 256 possible groups of 5

Once forming that first group of 5, the remaining 5 players would all be placed on the second team by default.

The 'twist' is that the two teams of 5 can "show up" in either order:

For example, if we call the two teams of 5 players: A,B,C,D,E and F,G,H,I,J

ABCDE vs. FGHIJ

is the SAME match-up as....

FGHIJ vs. ABCDE

So we are NOT allowed to count that matchup twice. This means we have to divide the 256 by 2.

[Reveal] Spoiler:
C

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

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22 Jun 2016, 06:01
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Jill is dividing her ten-person class into two teams of eq   [#permalink] 22 Jun 2016, 06:01
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