



 
Steve, The table you have is from Victoria's response to Evan. I am going to add a row to the table to indicate the golfers. I am naming the golfers S (that's you) and A, B, C, D, E, F and G.
Each day the golfers are divided into two foursomes called 0 and 1. Each player determines which foursome he or she belongs to by reading down his or her column. Steve, reading down your column you see all zeros so you are in foursome 0 every day. Look now, for example at golfer D. His column reads 1, 0, 0, 1, 1, 1, 0. Thus on Day 0 he is not in your foursome, on Day 1 and Day 2 he is in your foursome. On Days 3, 4, and 5 he is not in your foursome and on Day 6 he is in your foursome. I hope this explains Victoria's table. Write back if you need more assistance. Harley In March of 2015 we received a further question from Gord.
Gord, It turns out that there is only one schedule, up to rearrangement of the days or renumbering of the players. The latter is irrelevant. So your only hope is to shuffle the days. Unfortunately, there are $7! = 5040$ possible ways to do that. If you have a friend who can program, then s/he could write a small program to look at all of the possible orderings of the days, count the number of “three in a rows”, and keep the best schedule. Good luck.  


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