20 golfers that play in groups of four each day. How many days will it take so that each golfer plays with all of the other golfers at least once?
in Word Problem Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

If the golfers (A to T) were to play in pairs each golfer the pattern of games would be:

AB ...AT (19), BC...BT (18), CD...CT (17), ...,PQ PR PS PT (4), QR QS QT (3), RS RT (2), ST (1).

The sum of the numbers 1 to 19=190.

There are 190 unique combinations of players.

If two pairs are combined we know that all the possible combinations will be present, and there will be duplications within the groups, but every player will have played at least once with every other player. Those combinations of pairs result in half as many games=190/2=95. However, this appears to be the maximum number of games. 

However, it has been assumed that the exercise of combining two pairs doesn’t include the same player twice. For example, GH can’t be combined with HK. 

Below is a solution containing 58 games (this may not be the minimum).

by Top Rated User (680k points)

Related questions

1 answer
asked Apr 27, 2013 in Algebra 2 Answers by anonymous | 191 views
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
83,678 questions
88,568 answers
1,985 comments
5,801 users