The second stanza of a sestina, as we saw last week, rearranges the endings of the six verses from ABCDEF to FAEBDC. If we apply the same criteria to go from the second to the third, from the third to the fourth and so on, we obtain the sequence:

ABCDEF, FAEBDC, CFDABE, ECBFAD, DEACFB, BDFECA.

And if we change the traditional capital letters that in poetic notation indicate the endings of the verses of major art for numbers and we arrange the sequences corresponding to the successive stanzas vertically, we obtain the following scheme:

1 6 3 5 4 2

2 1 6 3 5 4

3 5 4 2 1 6

4 2 1 6 3 5

5 4 2 1 6 3

6 3 5 4 2 1

There are no repeated figures in any row or column, so the sestina scheme is like a reduced sudoku, with the numbers 1 to 6 instead of 1 to 9. Although, for mathematicians, before a sudoku it is a Latin square. And this time poetry could have preceded mathematics, since the first sestinas were composed in the 12th century by the Occitan troubadour Arnaut Daniel, while the first Latin squares (named by Euler much later) of which there is news are the *wafq majazi* from an Arabic manuscript of the 13th century.

### Combinatorial design theory

As for the problem that had been left pending (finding out how many ways seven elements can be grouped into seven groups of three elements, if they have to appear in the same number of groups and two by two only in one group), here is the solution provided by Ignacio Alonso:

“Each element will be in three trios. To P. For example, the 7, the associated trio that contains the 6 will be with the duos 65, 64… 61 (5 possible). If the first is 765, the second associated that contains the 4 could be 743, 742 or 741 (3 possibilities) and the third associated with 765 and 743 can only be 72. Total, 5 × 3 = 15 groups of three possible trios that contain the 7. The four remaining trios without the 7, associated with a group of these 15, be it 765, 743, 721, contain twice the 65, 64… 61. With the 6 the possible ones are 642, 631 or 641 , 632 (2 possibilities), for each of these two, e.g. 642, 631, only a single associate, 541, 532, to complete this group of four trios, then 15 × 2 = 30 will be the groups of 7 possible trios.”

As we saw, this problem could be considered a simplified version of Kirkman's classic “schoolgirl problem”, of which there are only seven non-isomorphic solutions (that is, of non-equivalent structures). But if we include isomorphic solutions, the number increases considerably (can you calculate it?).

These problems – and also the Latin squares – have to do with the so-called “theory of combinatorial design”, developed from the pioneering contributions of Leonard Euler, Thomas Kirkman, Jacob Steiner, Édouard Lucas and other great mathematicians of the 18th and 18th centuries. XIX; theory that, by the way, owes not a little to recreational mathematics. But that's another article.

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