222 kibkman's school-girls problem [ch. X 



other days. I conjecture that similar methods are applicable to 

 corresponding problems about quintets, sextets, &c. 



Bridge Problem. Another analogous question is where we deal 

 with arrangements in pairs instead of triplets. One problem of 

 this kind is to arrange 4m members of a bridge club for 4m — 1 

 rubbers so that (i) no two members shall play together as 

 partners more than once, and (ii) each member shall meet every 

 other member as opponent twice. The general theory has been 

 discussed by E. H. Moore, 0. Eckenstein and G. A. Miller. A 

 typical method for obtaining cyclic solutions is as follows. Denote 

 the members by a point k at the centre of a circle and by 4m — 1 

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

 We can join the points 2, 3, 4, ... by chords, and these chords 

 with (k. 1) give possible partners at the m tables in the first 

 rubber. A one-step cyclical permutation of the numbers will 

 give the arrangements for the other rubbers if, in the initial 

 arrangement, (i) the lengths of the chords representing every 

 pair of partners are unequal and thus appear only once, and (ii) 

 the lengths of the chords representing every pair of opponents 

 appear only twice. Since the chords representing pairs of 

 partners are unequal, their lengths are uniquely determined, but 

 the selection of the chords is partly empirical. Solutions for 

 many values of m have been given. In the following examples 

 for m = 2, 3, 4, 1 give an arrangement of the card tables for the 

 first rubber : the arrangements for the subsequent rubbers being 

 thence obtained by one-step cyclical permutations of the 

 numbers. If m = 2, such an initial arrangement is (k. 1 against 



5. 6) and (2. 4 against 3. 7). If m = 3, one such initial arrange- 

 ment is (k. 1 against 5. 6), (2. 11 against 3. 9) and (4. 8 against 

 7. 10). If m = 4, one such initial arrangement is (k. 1 against 



6. 11), (2. 3 against 5. 9), (4. 12 against 13. 15), and (7. 10 

 against 8. 14). There are also solutions by other methods. 



Sylvester's Corollary. To the original theorem J. J. Sylvester 

 added the corollary that the school of 15 girls could walk out in 

 triplets on 91 days until every possible triplet had walked abreast 

 once, and he published a solution in 1861. 



The generalized problem of finding the number of ways in 



