Question

I am trying to build an algorithm which chooses a combination of 4 horses from a number of races.

So say we have 10 races, each with 8 horses running: I want all combinations of 4 horses from these races, with the condition that the 4 horses have to come from separate races (no two horses in the same combination can come from the same race).

Is there a name for this problem (and existing algorithm)?

I assume it’ll just be a series of loops but I haven’t had my coffee today…
Cheers

EDIT: I should say for each combination I perform a fairly lengthy function on the combination, so I don’t want to repeat combinations I’ve already done.

EDIT: Any better way than this?

            n = number of races;
            for ( int i = 0; i < n; i++ ) {
                for ( int j = i + 1; j < n; j++ ) {
                    for ( int k = j + 1; k < n; k++ ) {
                        for (int l = k + 1; l < n; l++) {
                            //for each combination of 4 separate sets
                            for(int p = 0; p < races.get(i).getHorses().size(); p++){
                                for(int q = 0; q < races.get(j).getHorses().size(); q++){
                                    for(int r = 0; r < races.get(k).getHorses().size(); r++){
                                        for(int s = 0; s < races.get(l).getHorses().size(); s++){
                                            //each combination of 4 horses
                                            races.get(i).getHorses().get(p)
                                            races.get(j).getHorses().get(q)
                                            races.get(k).getHorses().get(r)
                                            races.get(l).getHorses().get(s)
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }

Answer 1

Basically you have two problems. One to choose 4 races out of the 10 total (order doesn’t matter), which is 10! / (6! * 4!) or 210 combinations. Then, for each combination of 4 races you can choose any of the 8 horses in each race, which is 8^4 or 4096.