208 kirkman's school-girls problem [ch. X 



one-step cyclical permutations of the suffixes, the arrangements for 21 days can 

 be got. Each of the 10 triplets, consisting of the 3 triplets (c20. 611. al2), 

 (cl6. 613, al8), (cl7. 62. a6), the 6 analogous triplets, and the triplet (ol. 61. cl), 

 gives, by a one-step cyclical permutation of the suffixes, an arrangement for 

 one of the remaining 10 days. 



For 81 girls, an arrangement for the first day consists of the 9 triplets 

 (o5. o7. 08), (o3. alO. ol4), (oil. o21. o26), (o4. al2. a25), (a9. al8. 627), 

 (o22. 620. cl9), (o24. 617. cl3), (ol6. 66. cl), (a23. 615. c2), and the 18 analogous 

 triplets. From these, by cyclical permutations of the suffixes, the arrangements 

 for 27 days can be got. Each of the 13 triplets consisting of the 4 triplets 

 (c22. 620. ol9), (c24. 617. al3), (cl6. 66. ol), (c23. 615. o2), the 8 analogous 

 triplets, and the triplet (al. 61. cl), gives, by a cyclical permutation of the 

 suffixes, an arrangement for one of the remaining 13 days. 



For 99 girls, an arrangement for the first day consists of the 11 triplets 

 (ol. o3. olO), (o2. a6. o20), (o4. ol2. o7), (08. a24. ol4), (ol6. ol5. a28), 

 (oil. o22. 633), (o32. 630. c23), (o31. 627. cl3), (o29. 521. c26), (o25. 59. cl9), 

 (ol7. 618. cS), and the 22 analogous triplets. From these, by cyclical permuta- 

 tions of the suffixes, the arrangements for 33 days can be got. Each of the 16 

 triplets consisting of the 5 triplets (c32. 630. o23), (c31. 527. al3), (c29. 621. a26), 

 (c25. 69. ol9), (cl7. 618. o5), the 10 analogous triplets, and the triplet 

 (ol. 61. cl), gives, by a cyclical permutation of the suffixes, an arrangement for 

 one of the remaining 16 days. 



For 27 girls, an arrangement of triplets for the first day is (1. 7. 10), (14.17.8), 

 (27. 6. 9), (13. 20. 25), (11. 23. 18), (12. 16. 24), (21. 2. 4), (15. 22. 26), (19. 3. 5). 

 From this, three-step cyclical permutations give arrangements for 9 days in all. 

 The first triplets on the remaining 4 days are (1. 2. 3), (1. 6. 20), (1. 11. 15), 

 (1. 17. 27), from each of which, by a three-step cyclical permutation, an arrange- 

 ment for one of those days is obtainable. 



Regular solutions in the numerical notation can also be obtained for all 

 values of n, except 9, where n is of the form 18m +9. I give 27 girls as 

 an instance. The first day's arrangement is (1. 2. 4), (10. 11. 13), (19. 20. 22), 

 (8. 15. 23), (17. 24. 5), (26. 6. 14), (3. 25. 9), (12. 7. 18), (21. 16. 27); from this, 

 arrangements for 9 days in all are obtained by one-step cyclical permutations. 

 . Each of the three triplets (1. 5. 15), (2. 6. 16), (3. 7. 17) gives an arrangement 

 for one day by a three-step cyclical permutation. Finally the triplet (1. 10. 19), 

 represented by an equilateral triangle, gives the arrangement on the last day by 

 a one-step cyclical permutation. 



Lastly, I consider those cases where n is of the form 18m + 15. As before, 

 the solution is divided into two parts. In the first part, we obtain an arrange- 

 ment of triplets for the first day, from which arrangements for 6m + 5 days are 

 obtained by three-Btep cyclical permutations. In the second part, we obtain the 

 first triplet on each of the remaining 3m + 2 days, from which the other triplets 

 on that day are obtained by three-step cyclical permutations. 



For 33 girls, an arrangement in the first part is (1. 13. 19), (23. 11. 5), 

 (3. 15. 21), (2. 4. 7), (12. 14. 17), (28. 30. 33), (6. 16. 25), (10. 20. 29), (8. 18. 27), 

 (24. 31. 32), (9. 22. 26). The triplets in the second part are (25. 32 33) 

 (26. 33. 1), (10. 23. 27), (11. 24. 28), (1. 12. 23). " 



