SOLUTION: In a round robin competition, everyone plays everyone else at least once. How many games would be played if there were six players? How many more games would be needed if three e

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Question 1163013: In a round robin competition, everyone plays everyone else at least once. How
many games would be played if there were six players? How many more games
would be needed if three extra players entered the competition?
Answer by ikleyn(52777) About Me  (Show Source):
You can put this solution on YOUR website!
.

To be correct, the problem should be formulated in DIFFERENT WAY: 

    In a round robin competition, everyone plays everyone else highlight%28cross%28at_least_once%29%29 exactly one time. 
    How many games would be played if there were six players? 

    How many more games would be needed if three extra players entered the competition?


Then the answer to the first question is   %286%2A5%29%2F2 = 3*5 = 15  games.


The answer to the second question is   %289%2A8%29%2F2 - 15 = 9*4 - 15 = 36 - 15 = 21  games   (the difference).

Solved.

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For n participants, the number of games between them is   %28n%2A%28n-1%29%29%2F2.