CH. X] KIRKMAN'S SCHOOL-GIKLS PROBLEM 221 



sixth group, k6, ai, 69, ol, dS, el, /2, g%, h5 ; seventh group, 

 hi, a6, 65, c9, d2, e4, /8, g3, Al ; eighth group, k8, al, 67, c6, 

 <29, e.3,/5, #1, M; ninth group, k9, al, 62, c3, d4, e5,/6, #7, A8. 

 The arrangements for the other eight days are got by shifting 

 the numbers attached to the letters a,b,...h (but not k) in 

 each group cyclically. Thus the first group on the third day 

 will be, Id, aB, 68, c7, d9, e4,/6, g2, h5. 



When n is composite no general method of attacking the 

 problem has been discovered, though solutions for various par- 

 ticular cases have been given. But when n is prime, we can 

 proceed thus. Denote the w 3 girls by the suffixed letters shown 



in the left half of the above table. Take this as giving the 

 arrangement on the first day. Then on the second day we may 

 take as an arrangement that shown in the right half of the 

 table. From the arrangement on the second day, the arrange- 

 ments for the other days are obtained by one-step cyclical 

 permutations of the suffixes of a, b, &c. ; the suffixes of k being 

 unaltered. 



Kirkmaris Problem in Quartets. The problem of arranging 

 4m girls, where m is of the form 3n + 1, in quartets to walk out 

 for (4m — 1)/3 days, so that no girl will walk with any of her 

 school-fellows in any quartet more than once has been attacked. 

 Methods similar to those given above are applicable, and solu- 

 tions for all cases where m does not exceed 49 have been written 

 out. One example will suffice: for 16 girls (i.e. when m = 4, 

 n = 1), arrangements in quartets for five days can be obtained 

 from the following arrangement for the first day, (k. 5. 10. 15), 

 (8. 2. 1. 4), (13. 7. 6. 9), (3. 12. 11. 14) ; from which three-step 

 cyclical permutations of the numbers give arrangements for the 



