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# I have 12 golfers playing over 4 days (once per day) ...

### Resolved Question:

I have 12 golfers playing over 4 days (once per day) in groups of four. Is there a way that every player can play with each other at least once?
Submitted: 8 years ago.
Category: General
Expert:  Scott replied 8 years ago.

HiCustomer and welcome to JA!

The answer is no. There is no way for all 12 players to play with every other player in just 4 rounds.

I'm not sure if this is a math problem or if you are actually having a golf event, so here is a detailed description of why that is the case:

• Divide the players into pairs. There are 66 (12*11 / 2), or 12 C 2 if you are into probability.
• Each group of 4 determines 6 pairs. (4*3 / 2), or 4 C 2.
• Therefore 3 groups of 4 is 18 pairs.
• BUT, if you randomly assign people to the first round of play, there is no way of assigning them to the next round without repeating at least 3 pairs (you can see the "optimal" schedule below if you don't believe me).
• Therefore, for each of the next rounds, we have 3 fewer choices.
• So the total number of pairs we can have is 18+15+12+9 = 54
• 54 matched pairs is less than the 66 that we originally calculated exist, so this is impossible to get all the pairs together.

Here is the "most" optimal schedule for players A1, A2, A3, A4, B1, B2, B3, B4, C1, C2, C3, and C4:

First day:

(A1,A2,A3,A4)
(B1,B2,B3,B4)
(C1,C2,C3,C4)

Day 2:

(A1,B1,C1,A2)
(B2,C2,A3,B3)
(C3,A4,B4,C4)

Day 3:

(A1,B2,C3,A4)
(B1,C2,A3,B4)
(C1,A2,B3,C4)

Day 4:

(A1,C2,A3,C4)
(B1,A2,B2,C3)
(C1,B3,B4,A4)

Refs:

http://mathcentral.uregina.ca/QQ/database/QQ.09.02/dale2.html

I hope this helps. Let me know if you have any questions,

Scott

Category: General
Satisfied Customers: 17710
Experience: MIT Graduate, United States Traveler, Information Finder

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