263.—­KING ARTHUR’S KNIGHTS.

On the second evening King Arthur arranged the knights and himself in the following order round the table:  A, F, B, D, G, E, C. On the third evening they sat thus, A, E, B, G, C, F, D. He thus had B next but one to him on both occasions (the nearest possible), and G was the third from him at both sittings (the furthest position possible).  No other way of sitting the knights would have been so satisfactory.

264.—­THE CITY LUNCHEONS.

The men may be grouped as follows, where each line represents a day and each column a table:—­

    AB CD EF GH IJ KL
    AE DL GK FI CB HJ
    AG LJ FH KC DE IB
    AF JB KI HD LG CE
    AK BE HC IL JF DG
    AH EG ID CJ BK LF
    AI GF CL DB EH JK
    AC FK DJ LE GI BH
    AD KH LB JG FC EI
    AL HI JE BF KD GC
    AJ IC BG EK HL FD

Note that in every column (except in the case of the A’s) all the letters descend cyclically in the same order, B, E, G, F, up to J, which is followed by B.

265.—­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.

    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.

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.

267.—­THE WRONG HATS.

The number of different ways in which eight persons, with eight hats, can each take the wrong hat, is 14,833.