0

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..

Community
  • 1
Michael Crook
  • 115

2 Answers2

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

David
  • 82,662
  • You got the correct answer (made a nifty program of which I will link at a later time which got this answer via brute force (I am writing a short blog post on this) but as I have been out of mathematics for a while, I am more interested in why you used (10 OVER 5) and how that turned into 10! OVER 5!5! which further simplified to 10.9.8.7.6.5 OVER 5.4.3.2.1 Do you have any resources that link to this mathematics? – Michael Crook Nov 12 '14 at 00:10
  • Here is a nice simple explanation. – David Nov 12 '14 at 00:23
  • Anyone interested: see AlexR comments, Binomial coefficients – Michael Crook Nov 12 '14 at 00:23
  • Answer is perfectly valid, unfortunatly so is AlexR's which was in a fraction of time before yours. Thanks a bunch! – Michael Crook Nov 12 '14 at 00:25
2

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 $$\frac1{M!} \prod_{j=0}^{M-1} \binom{N-jM}{k}$$

AlexR
  • 24,905
  • You got the correct answer (made a nifty program of which I will link at a later time which got this answer via brute force (I am writing a short blog post on this) but as I have been out of mathematics for a while, I am more interested in why you used (10 OVER 5) and how that turned into 10.9.8.7.6.5 OVER 5.4.3.2.1 Do you have any resources that link to this mathematics? – Michael Crook Nov 12 '14 at 00:09
  • 1
    @MichaelCrook The binomial coefficients are pretty standard. They are defined via $\binom nk := \frac{n!}{(n-k)!k!} = \frac1{k!} \prod_{j=0}^{k-1} n-k$ – AlexR Nov 12 '14 at 00:13
  • Ahhh and you got to 10.9.8.... because one of the 5!'s councelled out the 5.4.3.2.1 Awesome, makes perfect sense

    Thanks

     – Michael Crook Nov 12 '14 at 00:22
  • @MichaelCrook Yep. If anything remained unclear, feel free to ask for clarification ;) – AlexR Nov 12 '14 at 00:23
  • I don't yet understand the 2 other lines of math, but I plan to read more into them after work. Thanks, marking as answer as first to reply with correct answer. – Michael Crook Nov 12 '14 at 00:24
  • 1
    @MichaelCrook You may want to refer to this for the notational conventions used. – AlexR Nov 12 '14 at 00:25
  • 1
    Thanks, just did another 7 people in teams of 5 (dis-reguard the left over 2) got 21 in my program and 21 via the math so awesome, thanks. – Michael Crook Nov 12 '14 at 00:29