20 players are to be divided into two 10-man teams. In how many ways can that be done?

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- Jan 11th 2010, 12:03 PMkturfFundamental Counting Principles
20 players are to be divided into two 10-man teams. In how many ways can that be done?

- Jan 11th 2010, 01:52 PMDrexel28
- Jan 12th 2010, 05:17 AMkturfNot sure but ...
Here is a thought:

20!/10!(20-10)! - Jan 12th 2010, 06:32 AMPlato
- Jan 12th 2010, 06:39 AMOloria
I see it so. The sequence is not important, so

$\displaystyle {20 \choose 10}$ - Jan 12th 2010, 06:42 AMOloria
- Jan 12th 2010, 06:44 AMPlato
- Jan 12th 2010, 06:50 AMOloria
I noticed it on my own before. I read too quickly. But I asked you about something another...

- Jan 12th 2010, 07:03 AMPlato
Say we have a prep school class of fifteen boys.

We partition them into three study groups of five each to study: biology, mathematics, and geography.

This can be done in $\displaystyle \binom{15}{5}\binom{10}{5} \binom{5}{5}=\frac{15!}{(5!)^3} $ ways.

That is an ordered partition because names of the groups as well as the content of the of the group matters.

NOW, We partition them into three teams of five each to play.

This can be done in $\displaystyle \frac{15!}{(5!)^3(3!)} $ ways.

That is an unordered partition because the teams are not named so only the content of the of the team matters. - Jan 12th 2010, 07:19 AMkturfOK - that helps.
I think the answer is:

20!/(10!)^2(2!) - Jan 12th 2010, 07:26 AMPlato