[ 59 ] 



VII. On a Tactical Theorem relating to the Triads of Fifteen 

 Things. By A. Cayley, Esq.* 



THE school-girl problem may be stated as follows : — " With 

 15 things to form 35 triads, involving all the 105 duads, 

 and such that they can be divided into 7 systems, each of 5 

 triads containing all the 15 things." A more simple problem is, 

 "With 15 things, to form 35 triads involving all the 105 duads." 

 In the solution which I formerly gave of the school-girl pro- 

 blem (Phil. Mag. vol. xxxvii. 1850), and which maybe presented 

 in the form 



abc 

 acle 



rfy 



bdf 

 bge 

 cdg 



cef 



a 



b 



c 



d 



e 



/ 



9 









35 



17 



82 



64 





62 



84 







15 



37 





13 



57 



86 



42 







47 





16 





38 





25 



58 





23 



14 





67 





12 



78 







56 



34 





36 



45 





27 







18 



(viz. the things being a, b, c, d, e,f g, 1, 2, 3, 4, 5, 6, 7, 8, the 

 first pentad of triads is abc, dSo, e 17, f 82, g 64, and so for all the 

 seven pentads of triads), there is obviously a division of the 15 

 things into (7 + 8) things, viz. the 35 triads are composed 7 of 

 them each of 3 out of the 7 things, and the remaining 28 each of 

 1 out of the 7 things, and 2 out of the 8 things : or attending 

 only to the 8 things, there are 28 triads each of them containing 

 a duad of the 8 things, but there is no triad consisting of 3 

 of the 8 things. More briefly, we may say that in the system 

 there is an 8 without 3, that is, there are 8 things, such that no 

 triad of them occurs in the system. 



I believe, but am not sure, that in all the solutions which have 

 been given of the school-girl problem there is an 8 without 3. 



Now, considering the more simple problem, there are of course 

 solutions which have an 8 without 3 (since every solution of the 

 school-girl problem is a solution of the more simple problem) : 



* Communicated by the Author. 



