In the following solution each of the eleven lines represents a sitting, each column a table, and each pair of letters a pair of partners.
| A B — I L | E J — G K | F H — C D |
| A C — J B | F K — H L | G I — D E |
| A D — K C | G L — I B | H J — E F |
| A E — L D | H B — J C | I K — F G |
| A F — B E | I C — K D | J L — G H |
| A G — C F | J D — L E | K B — H I |
| A H — D G | K E — B F | L C — I J |
| A I — E H | L F — C G | B D — J K |
| A J — F I | B G — D H | C E — K L |
| A K — G J | C H — E I | D F — L B |
| A L — H K | D I — F J | E G — B C |
It will be seen that the letters B, C, D ...L descend cyclically. The solution given above is absolutely perfect in all respects. It will be found that every player has every other player once as his partner and twice as his opponent.
[266.—A TENNIS TOURNAMENT.—solution]
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 against D b | E a against B d |
| 3rd Day | A b against E d | B a against 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.