1. ## Fundamental Counting Principles

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

2. Originally Posted by kturf
20 players are to be divided into two 10-man teams. In how many ways can that be done?
What do you think?

3. ## Not sure but ...

Here is a thought:
20!/10!(20-10)!

4. Originally Posted by kturf
20 players are to be divided into two 10-man teams. In how many ways can that be done?
These are an unordered partitions.
We can divide thirty people in five groups of six each in $\displaystyle \frac{30!}{(6!)^5(5!)}$ways.

5. I see it so. The sequence is not important, so
$\displaystyle {20 \choose 10}$

6. Originally Posted by Plato
These are an unordered partitions.
We can divide thirty people in five groups of six each in $\displaystyle \frac{30!}{(6!)^5(5!)}$ways.

Could you show why?

7. Originally Posted by Oloria
I see it so. The sequence is not important, so a group: a, b is the same as a group: b, a.
$\displaystyle {10 \choose 2}{8 \choose 2}{6 \choose 2}{4 \choose 2}{2 \choose 2}$
There are only two teams of ten each. The teams are un-named.

8. I noticed it on my own before. I read too quickly. But I asked you about something another...

9. Originally Posted by Oloria
Could you show why?
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.