“I have been given to understand,” he added, “that some of you fellows are very fond of rowing, so I propose on this occasion to provide you with this recreation, and at the same time give you an amusing little puzzle to solve.  During the seven days that you are at Herne Bay every one of you will go out every day at the same time for a row, but there must always be three men in a boat and no more.  No two men may ever go out in a boat together more than once, and no man is allowed to go out twice in the same boat.  If you can manage to do this, and use as few different boats as possible, you may charge the firm with the expense.”

One of the men tells me that the experience he has gained in such matters soon enabled him to work out the answer to the entire satisfaction of themselves and their employer.  But the amusing part of the thing is that they never really solved the little mystery.  I find their method to have been quite incorrect, and I think it will amuse my readers to discover how the men should have been placed in the boats.  As their names happen to have been Andrews, Baker, Carter, Danby, Edwards, Frith, Gay, Hart, Isaacs, Jackson, Kent, Lang, Mason, Napper, and Onslow, we can call them by their initials and write out the five groups for each of the seven days in the following simple way: 

                 1 2 3 4 5
    First Day:  (ABC) (DEF) (GHI) (JKL) (MNO).

The men within each pair of brackets are here seen to be in the same boat, and therefore A can never go out with B or with C again, and C can never go out again with B. The same applies to the other four boats.  The figures show the number on the boat, so that A, B, or C, for example, can never go out in boat No. 1 again.

270.—­THE GLASS BALLS.

A number of clever marksmen were staying at a country house, and the host, to provide a little amusement, suspended strings of glass balls, as shown in the illustration, to be fired at.  After they had all put their skill to a sufficient test, somebody asked the following question:  “What is the total number of different ways in which these sixteen balls may be broken, if we must always break the lowest ball that remains on any string?” Thus, one way would be to break all the four balls on each string in succession, taking the strings from left to right.  Another would be to break all the fourth balls on the four strings first, then break the three remaining on the first string, then take the balls on the three other strings alternately from right to left, and so on.  There is such a vast number of different ways (since every little variation of order makes a different way) that one is apt to be at first impressed by the great difficulty of the problem.  Yet it is really quite simple when once you have hit on the proper method of attacking it.  How many different ways are there?

[Illustration]

271.—­FIFTEEN LETTER PUZZLE.