y
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 reall