Six managers at a company all enjoy golf. Each Saturday, four of the six get together for 18 holes of golf. They have decided to set up a schedule so that the same foursome does not play twice before all possible foursomes have played. The number of weekends that will pass before the same group would play twice is 15.

Question

Quizplus · Accepted Answer

The number of unique combinations of 4 managers out of 6 is calculated using the combination formula C(n, k) = n! / (k!(n-k)!), where n is the total number of managers and k is the number of managers per group. C(6, 4) = 15, so it will take 15 weekends for all possible foursomes to play.

Six Managers at a Company All Enjoy Golf