CH. x] kirkman's school-gikls problem 219 



number of ihe first set, the second number of the second set, 

 and the last number of the third set. In every other arrange- 

 ment the first triplet is formed from the first triplet in the next 

 proceeding arrangement, by adding unity to the number of the 

 second set, subtracting unity from the number of the third set, 

 and leaving the number of the first set unchanged. In every 

 arrangement the second and all subsequent triplets are formed 

 cyclically from the next preceding triplet. 



The arrangements thus made of the numbers in all the 

 groups of all the collections will give arrangements of all the 

 numbers for (Sim - l)/2 days, and will provide a solution of 

 the problem. 



Harison's Theorem provides solutions, alternative to those 

 given above, for the cases when n = 27, 45, 63, 75, 81, 99. 

 Also we can, by it, from the solutions already given for all 

 values of n less than 100, at once deduce solutions for the 

 cases when n= 105, 117, 135, 147, 153, 165, 171, 189, 195, &c. 



Extension to n* Girls. Peirce suggested the corresponding 

 problem of arranging n 2 girls in n groups, each group containing 

 n girls, on n + 1 days so that no two girls will be together in 

 a group on more than one day. We may conveniently represent 

 the girls by a point k at the centre of a circle and n % — 1 equi- 

 distant points, numbered 1, 2, 3, . . . , on the circumference. 



When n = 2, we may arrange initially the 4 points in two 

 pairs, one pair consisting of k and one of the points, say, 3, 

 and the other pair of the remaining points (1. 2). These two 

 pairs give the arrangement for the first day. From them, 

 the solution for the other days is obtained by one-step cyclical 

 permutations. 



When n = 3, we may arrange initially the 9 points in three 

 triplets, namely, k and the ends of a diameter (k. 4. 8); a 

 triangle (1. 2. 7) ; and the similar triangle (5. 6. 3) obtained by 

 a four-step cyclical permutation. These three triplets give an 

 arrangement for the first day. From them the solutions for 

 the other days are obtained by one-step cyclical permutations. 



When n = 4, we may arrange initially the 16 points in four 

 quartets, namely, k and three equidistant points, (k. 5. 10. 15); 



