| 1 | = | 0 |
| 2 | = | 1 |
| 3 | = | 2 |
| 4 | = | 9 |
| 5 | = | 44 |
| 6 | = | 265 |
| 7 | = | 1,854 |
| 8 | = | 14,833 |
To get these numbers, multiply successively by 2, 3, 4, 5, etc. When the multiplier is even, add 1; when odd, deduct 1. Thus, 3×1—1=2; 4×2+1=9 5×9-1=44; and so on. Or you can multiply the sum of the number of ways for and persons by , and so get the solution for persons. Thus, 4(2+9)=44; 5(9 +44)=265; and so on.
268.—THE PEAL OF BELLS.
The bells should be rung as follows:—
| 1 | 2 | 3 | 4 | 3 | 1 | 2 | 4 | 2 | 3 | 1 | 4 |
| 2 | 1 | 4 | 3 | 1 | 3 | 4 | 2 | 3 | 2 | 4 | 1 |
| 2 | 4 | 1 | 3 | 1 | 4 | 3 | 2 | 3 | 4 | 2 | 1 |
| 4 | 2 | 3 | 1 | 4 | 1 | 2 | 3 | 4 | 3 | 1 | 2 |
| 4 | 3 | 2 | 1 | 4 | 2 | 1 | 3 | 4 | 1 | 3 | 2 |
| 3 | 4 | 1 | 2 | 2 | 4 | 3 | 1 | 1 | 4 | 2 | 3 |
| 3 | 1 | 4 | 2 | 2 | 3 | 4 | 1 | 1 | 2 | 4 | 3 |
| 1 | 3 | 2 | 4 | 3 | 2 | 1 | 4 | 2 | 1 | 3 | 4 |
I have constructed peals for five and six bells respectively, and a solution is possible for any number of bells under the conditions previously stated.
269.—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 4557. 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) | (DEF) | (GHI) | (JKL) | (MNO) |
| 8 | 6 | 7 | 9 | 10 | |
| 2nd Day. | (ADG) | (BKN) | (COL) | (JEI) | (MHF) |
| 3 | 5 | 4 | 2 | ||
| 3rd Day. | (AJM) | (BEH) | (CFI) | (DKO) | (GNL) |
| 7 | 6 | 8 | 9 | 1 | |
| 4th Day. | (AEK) | (COM) | (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 ), 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.
270.—THE GLASS BALLS.
There are, in all, sixteen balls to be broken, or sixteen places in the order of breaking. Call the four strings A, B, C, and D—order is here of no importance. The breaking of the balls on A may occupy any 4 out of these 16 places—that is, the combinations of 16 things, taken 4 together, will be 13×14×15×161×2×3×4=1,920 ways for A. In every one of these cases B may occupy any 4 out of the remaining 12 places, making 9×10×11×121×2×3×4=495 ways. Thus 1920×495=950,400 different placings are open to A and B. But for every one of these cases C may occupy 5×6×7×81×2×3×4=70 different places; so that 950,400×70=66,528,000 different placings are open to A, B, and C. In every one of these cases, D has no choice but to take the four places that remain. Therefore the correct answer is that the balls may be broken in 66,528,000 different ways under the conditions.
Readers should compare this problem with No. 345, "The Two Pawns," which they will then know how to solve for cases where there are three, four, or more pawns on the board.
