(related to Problem: Three Men In A Boat)
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 of 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.
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