A TENNIS TOURNAMENT.
(
Combination and Group Problems)
Four married couples played a "mixed double" tennis tournament, a man
and a lady always playing against a man and a lady. But no person ever
played with or against any other person more than once. Can you show how
they all could have played together in the two courts on three
successive days? This is a little puzzle of a quite practical kind, and
it is just perplexing enough to be interesting.
Answer:
Call the men A, B, D, E, and their wives a, b, d, e. Then they may play
as follows without any person ever playing twice with or against any
other person:--
First Court. Second Court.
1st Day | A d against B e | D a against E b
2nd Day | A e " D b | E a " B d
3rd Day | A b " E d | B a " D e
It will be seen that no man ever plays with or against his own wife--an
ideal arrangement. If the reader wants a hard puzzle, let him try to
arrange eight married couples (in four courts on seven days) under
exactly similar conditions. It can be done, but I leave the reader in
this case the pleasure of seeking the answer and the general solution.
