250 
12 triads not combined Avith 13a in the above sex triplets 
form Avith it six quintuplets thus: — 
*0 
CO 
r-H 
13a 25 
3a 1 23 
3al 25 
la3 25 
la3 25 
89a 20 
46a 27 
891 56 
401 5- 
4 63 58 
79350 
45 a o3 
59a 71 
241 63 
281 7a 
29o 81 
263 m 
2 / a 3 g 
20a 14 
571 38 
561 al 
503 16 
583 n9 
60a 93 
CO 
601 82 
79 1 42 
783 62 
403 42 ; 
where the subindices form circles of five duads, and every 
triad of a quintuplet has the same final figure. 
We have combined 13a Avith all the other 54 triads, and 
in the same way Ave can form 66 quintuplets and 55 sextuplets 
Avhich shall once and once only exhaust the duads of the 
55 triads. 
We interpret the triad 13a 2S as the substitution of the 
second order, 
(13ajj)=l 537284609 a, 
Avhich has 13a undisturbed, and Avhose four transpositions 
are the subindices 25, 47, 68, 90, of 13a, 251, 253, 25a. 
The triads of the sextuplet (147 al ) are found to be the 
didymous radicals of a substitution A« of the sixth order, 
Avhose third power is <?(/ — 147„>, Avhich is permutable Avith the 
six triads of (147,,!). The quintuplets are sets of didymous 
radicals of substitutions of the fifth order, having circles of 
five subindices. 
Thus every pair AB of the 55 triads is either 0, ; 0 :) 0 b or 
one of the 55 triads. It remains to examine the triplets 
ABC. 
We easily prove that every irreducible triplet can be 
written as A(BC) 6 , when (BC) 6 is of the sixth order. The 
conditions that A should make A(BC)« reducible arc those 
above given in the like case. And avc can demonstrate, 
either a priori, or by inspection of the quintuplets and sextu- 
plets, that the only values of A Avhich make 
(A(BC) e =) A (13a 2J *670 39 ) 6 
irreducible are the tAvelve folloAving: — 124 3( „ 30a X8 , 78a J5 , 46a 27 , 
157 38 , 2 i a 39 , 140 j 7 , 361 79, 1 / 9 24 , 39a4 8 , 13845, 54a 3u . 
