Combinatorial Aggregation. 295 



Make, for example, x — 3, one of the synthemes in question 

 out of the twelve corresponding to this value will be 

 A.C.E.G.B.D.Fx2.3.6.3.6.3.6. 

 Here 

 A.C.E.G.B.D.F = A.C C.E E.G G.B B.D D.F F.A 



2.3.6 

 = 2.4 1 

 + 4.6 

 + 6.8 

 + 8.10 

 + 10.2 



:A 2 .C 4 , 



A 4 .C 6 , 



and the product 



C 3 .E 7 E 6 .G 4 G 3 .B 7 B 6 .D 4 D 3 .F 7 F 6 .A 3 

 C 5 .E 9 , E 8 .G 6 G 5 .B 9 B 8 .D 6 D 6 .F 9 F 8 .A 5 , 

 &c. &c. &c. 



To prove the rule for the table of formation, it will be suffi- 

 cient to show that no two contiguous duads ever contain the 

 same or equivalent permutations ; the equation of equivalence 

 it will be remembered isr:s = r + 2/ + 27»:s + 2*+2»*. 

 Now, as regards the first and second terms, it is manifest that 



1 : x cannot be equivalent, either to 1 : x 1 nor to 2 : x t nor to 



2 . x' t where x' is any number differing from x. 



Similarly, as regards the last and first term, x : I cannot be 

 equivalent to x' : 1, nor to x : 2, nor to x' : 2 ; therefore there 

 is no danger as far as the first term is concerned, either as an- 

 tecedent or consequent. 



Again, it is clear that x : 2 x — 1 cannot interfere with 

 x 1 : 2 x\ nor m + x . 2 x with m + x* : 2 x* — 1 ; neither can 

 2x — 1 : x with 2 x' : x', no r 2 x : m + x with 2 at — 1 : m + x 1 . 



Again, if possible, let x\ 2 x — 1 = m + x' : 2x' — 1 ; 

 then m + x 1 — x = 2 i 



and 2x' — 2x — 2/, 

 .*. 2 m = 2 i, 

 or m = r, 



which is impossible, since + i is the difference between two 

 indices, each less than m. 



Similarly, m + x : 2% cannot = x':2 x', 



and vice versa with the terms changed. 



2 x : m + x cannot = 2 x' : x 1 , 



and 2x — 1 : x cannot =2^ — 1 .m + x', 



which proves the rule for the table of formation. 



So much for the bipartite duad synthemes. As regards the 

 unipartite synthemes little need be said, for every part may 



