Combinatorial Aggregation, 291 



But the rule of proceeding must be now sufficiently obvious ; 

 the modulus being 2p, we divide the elements into p classes; 



p 1 . . 



these may be arranged into *—— — distinct forms of cyclothe- 



m 



matic arrangement, and each of the cyclothemes taken in four 



p— 1 

 positions, thus giving 4 X *-—t — , i. e. 2p — 2 bipartite syn- 



themes, the whole number that can be formed to the given 

 modulus 2 p. 



I shall now proceed to the theory of bipartite synthemes to 

 the modulus 2 m x p, by which it is to be understood that we 

 have p, parts each containing 2 m terms, and p is at present 

 supposed to be a prime number; the total number of syn- 

 themes to the modulus 2mp being 2mp — 1, and 2 m — 1 of 

 these evidently being capable of being made unipartite; the 

 remainder, 2 m p — 2 m, i. e. (p — 1) 2 m, will be the number 

 of bipartites to be obtained * : 



V — 1 



2m.(p — 1) = *—- — x 4 m ; 



^— c — denotes the total number of cyclothemes to modulus p ; 



m 



4- m, as will be presently shown, the number of lines or syzy- 

 gies in the Table of position. 



To fix our ideas let the modulus be 4 x 3, and let A, B, C 

 be three parts : 



a l a i a z a^\ 



b x b 2 b 3 b 4 > their constituents respectively. 



C \ C 2 C 3 C 4j 



Give ajixed order to the constituents of each part, then each 

 of them may be taken in four positions; thus A may be 

 written 



a x a 2 a 3 a A 



a 2 a 3 a 4 a l 



a 3 a 2 a x a 2 



a 4 a x # 2 a 3 . 



Assume some particular position for each, as, for instance, 



a x b x c x 

 « 2 Z> 2 c 2 



a 3 °3 C 3 



a 4 b 4 Cq 

 and read off by coupling the first and third vertical places of 



* In general, if there be t parts of /& terms each, and /a sr be even, the 

 number of bipartite synthemes is Or— 1) ft, as is easily shown from dividing 

 the whole number of bipartite duads by the semi-modulus. 



U2 



