# Golf Problem

We are golfing for 4 days.

If I mark each player with a letter.

a,b,c,d,e,f,g,h,i,j,k,l.

I need to find a calulation so everyone plays with each other at least once.

Thank you for trying

Carlton2772

## 1 Answer

\[{12\choose 2}=\frac{12!}{10!2!}=66.\]

So there is going to be a total of $66=4\times 16+2$ games. So you can arrange 16 games for three days and 18 games for one day.

You can have 3 groups of 4 each day, so the 4 players in a group only play with each other. Then in the first day there will be

\[3\times{4\choose 2}=18\]

games. Then you arrange 16 games for the remaining 3 days.

Poincare

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