THREE MEN IN A BOAT.
(
Combination and Group Problems)
A certain generous London manufacturer gives his workmen every year a
week's holiday at the seaside at his own expense. One year fifteen of
his men paid a visit to Herne Bay. On the morning of their departure
from London they were addressed by their employer, who expressed the
hope that they would have a very pleasant time.
"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.
Answer:
If there were no conditions whatever, except that the men were all to go
out together, in threes, they could row in an immense number of
different ways. If the reader wishes to know how many, the number is
455^7. And with the condition that no two may ever be together more than
once, there are no fewer than 15,567,552,000 different solutions--that
is, different ways of arranging the men. With one solution before him,
the reader will realize why this must be, for although, as an example, A
must go out once with B and once with C, it does not necessarily follow
that he must go out with C on the same occasion that he goes with B. He
might take any other letter with him on that occasion, though the fact
of his taking other than B would have its effect on the arrangement of
the other triplets.
Of course only a certain number of all these arrangements are available
when we have that other condition of using the smallest possible number
of boats. As a matter of fact we need employ only ten different boats.
Here is one the arrangements:--
1 2 3 4 5
1st Day (ABC) (DBF) (GHI) (JKL) (MNO)
8 6 7 9 10
2nd Day (ADG) (BKN) (COL) (JEI) (MHF)
3 5 4 1 2
3rd Day (AJM) (BEH) (CFI) (DKO) (GNL)
7 6 8 9 1
4th Day (AEK) (CGM) (BOI) (DHL) (JNF)
4 5 3 10 2
5th Day (AHN) (CDJ) (BFL) (GEO) (MKI)
6 7 8 10 1
6th Day (AFO) (BGJ) (CKH) (DNI) (MEL)
5 4 3 9 2
7th Day (AIL) (BDM) (CEN) (GKF) (JHO)
It will be found that no two men ever go out twice together, and that no
man ever goes out twice in the same boat.
This is an extension of the well-known problem of the "Fifteen
Schoolgirls," by Kirkman. The original conditions were simply that
fifteen girls walked out on seven days in triplets without any girl ever
walking twice in a triplet with another girl. Attempts at a general
solution of this puzzle had exercised the ingenuity of mathematicians
since 1850, when the question was first propounded, until recently. In
1908 and the two following years I indicated (see _Educational Times
Reprints_, Vols. XIV., XV., and XVII.) that all our trouble had arisen
from a failure to discover that 15 is a special case (too small to enter
into the general law for all higher numbers of girls of the form 6n+3),
and showed what that general law is and how the groups should be posed
for any number of girls. I gave actual arrangements for numbers that had
previously baffled all attempts to manipulate, and the problem may now
be considered generally solved. Readers will find an excellent full
account of the puzzle in W.W. Rouse Ball's _Mathematical Recreations_,
5th edition.
Informational
