point M represents the Monk, the point I the Island, and the point Y the Monastery. Now the only direct ways from M to I are by the bridges a and b; the only direct ways from I to Y are by the bridges c and d; and there is a direct way from M to Y by the bridge e. Now, what we have to do is to coimt all the routes that will lead from M to Y, passing over all the bridges, a, b, c, d, and e once and once only. With the simple diagram under the eye it is quite easy, without any elaborate rule, to count these routes methodically. Thus, starting from a, b, we find there are only two ways of completing the route; with a, c, there are only two routes; with a, d, only two routes; and so on. It will be foimd that there are sixteen such routes in all, as in the following list: —
| a | b | e | c | d | b | c | d | a | e |
| a | b | e | d | c | b | c | e | a | d |
| a | c | d | b | e | b | d | c | a | e |
| a | c | e | b | d | b | d | e | a | c |
| a | d | c | b | e | e | c | a | b | d |
| a | d | e | b | c | e | c | b | a | d |
| b | a | e | c | d | e | d | a | b | c |
| b | a | e | d | c | e | d | b | a | c |
If the reader will transfer the letters indicating the bridges from the diagram to the corresponding bridges in the original illustration, everything will be quite obvious.
262.—THOSE FIFTEEN SHEEP.
If we read the exact words of the writer in the cyclopaedia, we find that we are not told that the pens were all necessarily empty! In fact, if the reader will refer back to the illustration, he will see that one sheep is already in one of the pens. It was just at this point that the wily farmer said to me, "Now I'm going to start placing the fifteen sheep." He thereupon proceeded to drive three from his flock into the already occupied pen, and then placed four sheep in each of the other three pens. "There," says he, "you have seen me place fifteen sheep in four pens so that there shall be the same number of sheep in every pen." I was, of course, forced to admit that he was perfectly correct, according to the exact wording of the question.
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.
| AB—IL | EJ—G K | FH—CD |
| AC—JB | FK—HL | GI—DE |
| AD—KC | GL—IB | HJ—EF |
| AE—LD | HB—JC | IK—FG |
| AF—BE | IC—KD | JL—GH |
| AG—CF | JD—LE | KB—HI |
| AH—DG | KE—BF | LC—IJ |
| AI—EH | LF—CG | BD—JK |
| AJ—FI | BG—DH | CE—KL |
| AK—GJ | CH—EI | DF—LB |
| AL—HK | DI-FJ | EG—BC |
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 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.
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.
Here are the successive solutions for any number of persons from one to eight:—
