Mr. W. Spottiswoode on a Problem in Combinatorial Analysis. 351 



tion 13, 14, 15, the first determinant will be 



12 U 10 9 



] .5 1.6 1.7 1.8 



11 12 9 10 



2.5 2.6 2.7 2.8 



10 9 12 11 



3.5 3.6 3.7 3.8 



9 10 11 12 



4.5 4.6 4.7 4.8; 

 and omitting the superior numbers, the admissible terms will be 



(1.5) (2.7) (3.8) (4.6), (1.5) (2.8) (3.6) (4.7), 



(1.6) (2.7) (3.5) (4.8), (1.6) (2.8) (3.7) (4.5), 



(1.7) (2.6) (3.8) (4.5), (1.7) (2.5) (3.6) (4.8), 



(1.8) (2.5) (3.7) (4.6), (1.8) (2.6) (3.5) (4.7),; 

 repeating the process for the remaining 6 combinations in the 

 triangle (2), i. e. for the remaining 6 days, there will result 8 

 terms for each day, from each of which groups one must be selected 

 in such a manner that no combinations recur. The result is 



15.1.2 14.1.3 13.1.4 12.1.5 11.1.6 10.1.7 9.1.8 

 14.6.8 13.6.7 15.5.6 15.7.8 12.3.7 14.2.4 15.3.4 

 11.4.7 12.4.8 11.3.8 13.2.3 10.2.8 13.5.8 14.5.7 

 10.3.5 11.2.5 9.2.7 10.4.6 9.4.5 9.3.6 12.2.6 

 9.12.13 15.10.9 14.12.10 14.11.9 15.14.13 15.12.11 13.11.10 

 which, on writing 



l = a, 2 = b, 3 = c, 4=/, 5 = d, 0,-cj, 7 = e, 8 = h, 

 15 = i, 14=/', 13 = *, 12 = /, 11 ~m, 10 = rc, 9 = o, 

 will agree with the solution given by Mr. Cayley in this Maga- 

 zine, vol. xxxvii. p. 50. 



This may be exhibited under a rather more general point of 

 view, whereby all the tentative process is avoided, as follows. 

 The 7 determinants of the fourth degree above mentioned are 

 connected with the determinant 



