ch. x] kirkman's school-girls problem 205 



sides is a multiple of 3, must be used in the first part 

 of the solution. In the whole arrangement every possible 

 difference will occur n times, and, since any two assigned 

 numbers can occur together only once, each difference when 

 added to a number must start each time from a different 

 number. I will not go into further details as to how these 

 triangles are determined, but I think the above rules will be 

 clear if I apply them to one or two easy examples. 



For 9 girls, the possible differences are [1], [2], [3], [4], 

 each of which must be used three times in the construction of 

 four triangles the lengths of whose sides p, q, r are such that 

 p + q = r or p+q + r=9. One possible set of triads formed 

 from these numbers is [1, 2, 3], [1, 2, 3], [2, 3, 4], and [1, 4, 4]. 

 Every triangle with a side of the length [3] must appear in the 

 first part of the solution ; thus the triplets used in the first part 

 of the solution must be obtained from the first three of these 

 triads. Hence we obtain as an arrangement for the first day 

 the triplets (1. 3. 9), (2. 4. 7), (5. 6. 8). From this, three-step 

 cyclical permutations give arrangements for other two days. 

 The remaining triad [1, 4, 4] leads to a triplet (1. 2. 6) which, 

 by a three-step cyclical permutation, gives an arrangement for 

 the remaining day. 



If we use the suffix notation, an arrangement for the first 

 day is o^as, b^c^, Cia 2 c 3 . From this, by simple cyclical per- 

 mutations of the suffixes, we get arrangements for the second 

 and third days. Lastly, the triplet aJhCi gives, by cyclical 

 permutation of the suffixes, the arrangement for the fourth day, 

 namely, Oi&iCj, a 2 6 2 c a) a 3 b s c g . 



For 15 girls, the three-step process is inapplicable. The 

 explanation of this is that two triads are required in the 

 second part of the solution, and in neither of them may a 3 

 appear. The triads are to be formed from the differences 

 [1], [2], ... [7], each of which is to be used three times, and the 

 condition that in any particular triad only one 3 or one 6 may 

 appear necessitates that six of the triads shall involve a 3 or 

 a 6. Hence only one triad will be available for the second part 

 of the solution. 



