Number of unique Team parings given 10 players and 2 teams

Question

I yammer a wee bit too much, feel free to skip to TLDR unless you want more background as to why I care about this problem.

I was just thinking that it would be a fun to figure out the best 5 players in a team of 10 as this tournament only allows for 5 players.

Obviously, I could just go for an individually based analysis of each players merit, but I thought a competition of how many wins can each player get when playing on all possible team combinations would be cool!

As a programmer my first solution would be to simply create a small algorithm that got every possible combination and then sorted them by name and compared to find only the unique entries.. But that seems lazy.. I know the number of teams is probably going to be 60+ unqiue teams though so doing by hand is just asking for trouble (fairly sure maximum possible combinations of five is 5^10 according to stackoverflow before I find the unique ones) so I will probably have to rely on this bruteforce solve..

But it got me thinking... Surely there's a mathematical way to figure out the number of possible unique (order not mattering) teams given 5 players per team, 10 people and thus 2 teams.

TLDR:

I was wondering if anyone knows the proper way to solve this problem: You have 10 players and 2 teams. How many unique team parings can you make in which ordering doesn't matter.

I did some googling but I couldn't find any resources on this :(. A Link to an article/paper on the math behind this will entertain me enough, but if it's simple enough to be explained as an answer to this question directly with working... That would rock..

Answer 1

For two teams, it's the same as chosing $5$ players from the $10$ in any order and assigning them to Team A. The non-chosen players then form Team B. The number of ways to do so is $$\binom{10}5 = \frac{10\cdot9\cdot8\cdot7\cdot6}{5\cdot4\cdot3\cdot2\cdot 1} = 252$$

For $N$ players in $M$ teams of $k$ players each ($N=M\cdot k$) we select teams in order: $$\prod_{j=1}^M \binom{N-(j-1)\cdot M}{k}$$ And then divide by the ways to arrange the team names to give $$\frac1{M!} \prod_{j=0}^{M-1} \binom{N-jM}{k}$$

Answer 2

$\def\.{\times}$If I have understood the problem correctly, you have $10$ players and you want to select $5$ of them to form a team, with no player appearing twice on the same team (that is, $5$ different players) and order not important within a team.

The number of possible teams is $$C(10,5)=\binom{10}5=\frac{10!}{5!5!}=\frac{10\.9\.8\.7\.6}{5\.4\.3\.2\.1}=252\ .$$