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.