Business Idea and Stuck on The Math!!

[Jwhurricanes]

I have a business idea that requires some deep math beyond my limited skills. I am trying to determine how many unique combinations there are with 10 possible outcomes—how many 3 group combos and how many 5 group combos. Example: team 1 vs team 2
Team 3 vs team 4
5v 6
7 v 8
9 v 10

can’t have any teams playing each other as a part of the grouping—ex.1,2,3 or 1,2,4 because 1 and two play each other and there would be an automatic loser with that combo. This is the type of math problem I am trying to solve. Working on a number of different combinations—more teams, different combination of number of groups, etc. any smart people have a clue? Thank you in advance!! If you think this is simple please respond because I am willing to pay someone to help me through this.

 

[Zermelo]

Sorry, I don't quite get the problem. Could you please elaborate in more detail?

 

[Jwhurricanes]

Sorry, I don't quite get the problem. Could you please elaborate in more detail?

Team 1 (95) vs Team 2 (90)
Team 3 (88) vs Team 4 (85)
Team 5 (87) vs Team 6 (80)
Team 7 (75) vs Team 8 (73)
Team 9 (87) vs Team 10 (86)

5 games above with the final score in ( ). First of all, how many total 3 team combinations are there. Then, how many winning 3 team combinations are there. Remember that you can’t have teams that play each other in your combination of 3. Reason, is there would be a guaranteed loser in that grouping. So 1 and 2, 3,4 and so on through 9,10 can’t be in the same set of 3. Does that make sense? If not, feel free to contact me (phone# removed). If you think you can figure this out, I have a few more scenarios to figure out. Different amount of games with different amount of team combinations. I can pay you something for your time. Thank you so much and please, if it still doesn’t make sense, call me. Kind of hard to explain. Thanks again!!!!

 

[Zermelo]

For the first part, I guess you mean "In how many ways can I pick 3 teams out of these 10", aka 10 choose 3, which is 120. As for the second question, I guess you mean "In how many ways can I pick 3 teams out of the 5 winning teams", which is 5 choose 3 = 10.
I hope I figured out what you meant.

 

[Jwhurricanes]

Gotcha. The 120 would not have any teams that are playing each other in the “grouping”. So 1 and 2 would not be in any combination....same thing with 3,4...5,6 and so on? Is their a standard equation for figuring out both parts of my question? If I increased the grouping of three to five or if I increased the amount of games. Thank you in advance. Thank goodness for smart people like you??

 

[Zermelo]

Ooooh now I see what you meant. No, then 120 is not the answer. 120 is the number of all possible 3 element (team) subsets of the set with 10 teams. This is found using combinations without repetition, google “10 choose 3”. But your problem has a restriction, so we have to subtract the number of combinations that DO NOT satisfy this restriction, aka we need to find the “bad” combinations (for example, {1,2,3} is a bad combination, beacuse 1 and 2 played a game).
There are 5 pairs that played a game. Let’s assume that we chose a combination {1,2,x}, that is a bad combination because 1 and 2 played a game. The question is, how many of these combinations are there? Well, we can put any of the remaining elements as x, so there are 8 of those combinations. We repeat this process for the remaining bad pairs, and there are 5 of them, so, the number of all possible bad combinations is 5*8=40, so the final answer is 120-40=80.
This can be generalized, if you changed the grouping size from 3 to n, then the total number of combinations would be 10 choose n. The number of bad combinations would be 5*X, where X is the number of ways we can choose the remaining n-2 teams (which gets a little more complicated but nothing too much).

 

[Jwhurricanes]

You are my new favorite person. I think you nailed it. Not sure if I can communicate with you outside this forum but I would love to. Have a number of more scenarios that would love your help with. Wont let me give out number on here. Seriously want to thank you. Stranger helping a stranger is what it’s all about!!!!

 

[Zermelo]

You are my new favorite person. I think you nailed it. Not sure if I can communicate with you outside this forum but I would love to. Have a number of more scenarios that would love your help with. Wont let me give out number on here. Seriously want to thank you. Stranger helping a stranger is what it’s all about!!!!

No problem man, write me a private message and I will give you my email.