# Solution

(related to Problem: A Puzzle for Card-players)

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.

$$\begin{array}{ccc} A B — I L&E J — G K&F H — C D\\ A C — J B&F K — H L&G I — D E\\ A D — K C&G L — I B&H J — E F\\ A E — L D&H B — J C&I K — F G\\ A F — B E&I C — K D&J L — G H\\ A G — C F&J D — L E&K B — H I\\ A H — D G&K E — B F&L C — I J\\ A I — E H&L F — C G&B D — J K\\ A J — F I&B G — D H&C E — K L\\ A K — G J&C H — E I&D F — L B\\ A L — H K&D I — F J&E G — B C \end{array}$$

It will be seen that the letters $B, C, D,\ldots, L$ 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.

Thank you to the contributors under CC BY-SA 4.0!

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### References

#### Project Gutenberg

1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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