How can a program calculate combinations for three players without repetition?

[Oljeg]

Hi, I am trying to find the solution (formula) for doing the following:
For any number of players (divisible by three), I need to calculate a number of rounds (in my case 8) for three players to play each other without any repetition.
So the first round of etc. 21 players would be:
1,2,3 4,5,6, 7,8,9 10,11,12 13,14,15 16,17,18 19,20, 21 and I am looking for the combinations for the next 7 rounds so that no one plays the same player again. Eventually, I would like to design a program with number of players and rounds as an input variable.

Regards,
Oljeg

 

[phinds]

Have you studied finite math at all? There are formulas for permutations and combinations that you would likely find useful (combinations in your case, not permutations, since it doesn't matter who sits in what chair in your scenario)

 

[HallsofIvy]

Are you saying that each match involves three players? That's an odd kind of match! If each player must play each other player once, then each of the n players must be involved in (n- 1) matches. If each match involves 3 players, that would be a total of n(n- 1)/3 matches. For n= 21, this is 21(20)/3= 140 matches, 20 rounds of 7 matches per round.

 

[aikismos]

Yes, definitely a problem from combinatorics: (https://en.wikipedia.org/wiki/Combinatorics)

In general a combination of n elements taken k ways is given by:

$C(n,k) = \frac{n!}{(n-k)!}$

 

[micromass]

Yes, definitely a problem from combinatorics: (https://en.wikipedia.org/wiki/Combinatorics)

In general a combination of n elements taken k ways is given by:

$C(n,k) = \frac{n!}{(n-k)!}$

Wrong formula.

This is a problem of combinatorial designs. https://en.wikipedia.org/wiki/Combinatorial_design