Hi Ian,

Unfortunately you can't accomplish this.

Consider one particular golfer. To accomplish what you want he has to play with 27 other golfers. He plays with at most 3 different golfers each day for 7 days so that's 3 7 = 21 golfers he plays with, 6 short of the 27 required.

Penny

Ian wrote back

Hi, Thank you for your reply
Can you give me a schedule so that I can put them in foursomes over 7 days total 28 golfers using the best possible formula
I am trying to keep everybody happy
Thank you Ian

Ian,

If you go to the Quandaries and Queries page on Math Central and insert the word golf into the Quick Search field you will see some questions of golf schedules we have answered. In one of them

mathcentral.uregina.ca/QQ/database/QQ.02.06/jon3.html

you'll see a solution for 28 golfers over 3 days. If you look at what is happening in the columns from one day to the next (the first column is fixed, the second is moving up/around one notch, the third by 2 notches and the fourth by 3 notches) I think you can do quite well. Think of the columns like wheels that turn as in an odometer on your car (in the predigital days).

Penny