+108 Three schools have a chess tournament. Four players come from each school. Each player plays three games against each player from the other schools, and plays one game against each other player from his or her own school. How many games of chess are played?
+6250 There are 4C2 = 6 ways to select 2 schools.
Given those 2 schools there are 42 = 16 ways to pair up players 1 from each school
There are 3 games played by each of those 3 pairs.
This is a total of 6 x 16 x 3 = 288 games played against other schools.
There are 4C2 = 6 ways to select a pair from the 4 people at a given school.
There are 4 schools
They play 1 game each
This is a total of 6 x 4 x 1 = 24 games played against schoolmates
A total of 288 + 24 = 312 games
+108 Hey Rom, wouldn't it be 18 games against schoolmates? There are 3 schools not 4.
Rom: Have you seen this question?
https://web2.0calc.com/questions/thanking-in-advance-really-important
Is there a way of solving it, using permutations, combinations, factorials, partitions....etc. other than brute-force computer tabulation of each combination, as is done here in the solution appended to the question? I have tried and failed miserably!.
Thanks for looking at it.
