The case of 5 persons on 6 occasions may be solved as follows:—­

1 2 3 4 5 1 2 4 5 3 1 2 5 3 4 --------- 1 3 2 5 4 1 4 2 3 5 1 5 2 4 3

The case for 6 persons on 10 occasions is solved thus:—­

1 2 3 6 4 5 1 3 4 2 5 6 1 4 5 3 6 2 1 5 6 4 2 3 1 6 2 5 3 4 ----------- 1 2 4 5 6 3 1 3 5 6 2 4 1 4 6 2 3 5 1 5 2 3 4 6 1 6 3 4 5 2

It will now no longer be necessary to give the solutions in full, for reasons that I will explain.  It will be seen in the examples above that the 1 (and, in the case of 5 persons, also the 2) is repeated down the column.  Such a number I call a “repeater.”  The other numbers descend in cyclical order.  Thus, for 6 persons we get the cycle, 2, 3, 4, 5, 6, 2, and so on, in every column.  So it is only necessary to give the two lines 1 2 3 6 4 5 and 1 2 4 5 6 3, and denote the cycle and repeaters, to enable any one to write out the full solution straight away.  The reader may wonder why I do not start the last solution with the numbers in their natural order, 1 2 3 4 5 6.  If I did so the numbers in the descending cycle would not be in their natural order, and it is more convenient to have a regular cycle than to consider the order in the first line.

The difficult case of 7 persons on 15 occasions is solved as follows, and was given by me in The Canterbury Puzzles:—­

1 2 3 4 5 7 6 1 6 2 7 5 3 4 1 3 5 2 6 7 4 1 5 7 4 3 6 2 1 5 2 7 3 4 6

In this case the 1 is a repeater, and there are two separate cycles, 2, 3, 4, 2, and 5, 6, 7, 5.  We thus get five groups of three lines each, for a fourth line in any group will merely repeat the first line.

A solution for 8 persons on 21 occasions is as follows:—­

    1 8 6 3 4 5 2 7
    1 8 4 5 7 2 3 6
    1 8 2 7 3 6 4 5

The 1 is here a repeater, and the cycle 2, 3, 4, 5, 6, 7, 8.  Every one of the 3 groups will give 7 lines.

Here is my solution for 9 persons on 28 occasions:—­

2 1 9 7 4 5 6 3 8 2 9 5 1 6 8 3 4 7 2 9 3 1 8 4 7 5 6 2 9 1 5 6 4 7 8 3

There are here two repeaters, 1 and 2, and the cycle is 3, 4, 5, 6, 7, 8, 9.  We thus get 4 groups of 7 lines each.

The case of 10 persons on 36 occasions is solved as follows:—­

1 10 8 3 6 5 4 7 2 9 1 10 6 5 2 9 7 4 3 8 1 10 2 9 3 8 6 5 7 4 1 10 7 4 8 3 2 9 5 6

The repeater is 1, and the cycle, 2, 3, 4, 5, 6, 7, 8, 9, 10.  We here have 4 groups of 9 lines each.

My solution for 11 persons on 45 occasions is as follows:—­

2 11 9 4 7 6 5 1 8 3 10 2 1 11 7 6 3 10 8 5 4 9 2 11 10 3 9 4 8 5 1 7 6 2 11 5 8 1 3 10 6 7 9 4 2 11 1 10 3 4 9 6 7 5 8

There are two repeaters, 1 and 2, and the cycle is, 3, 4, 5,... 11.  We thus get 5 groups of 9 lines each.

The case of 12 persons on 55 occasions is solved thus:—­