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other letters of the set; we thus obtain all the substitutions which 
move every letter. Thus n = 5, we obtain the 44 substitutions for 
the letters abode , viz., these are 
(abode), &c., 24 symbols obtained by permuting in every way 
the four letters b , c, d , e. 
(< ab)(cde ), &c., 20 symbols corresponding to the 10 partitions 
ab, cde, and for each of them 2 arrangements 
such as cde, ced. 
And then if we reject those symbols which contain in any ( ) two 
consecutive letters, we have the substitutions which give the ar- 
rangements wherein the letter in the first place is not a or b, that 
in the second place not b or c, and so on. In particular n = 5, 
rejecting the substitutions which contain in any ( ), ab, be, cd, de, 
or ea , we have 13 substitutions, which may be thus arranged : — 
( acbed ) , (ac)(bed) , ( acebd ) , (adbec) , ( aedbc ) , 
( aedbc ) , ( bd)(aec ) , 
( acedb ) , (ce)(adb) , 
( aecbd ) , ( ad)(bec ) , 
( adeeb ) , ( be)(adc ) . 
Here in the first column performing on the symbol (acbed) the sub- 
stitution (abode'), we obtain ( bdeae ), = (aebdc), the second symbol ; 
and so again and again operating with ( abode ) we obtain the re- 
maining symbols of the column ; these are for this reason said to 
be of the same type. In like manner symbols of the second column 
are of the same type ; but the symbols in the remaining three 
columns are each of them a type by itself; viz., operating with 
(abode) upon (acebd) we obtain (bdace), = (acebd ) ; and the like as 
regards (adbec) and (aedbc) respectively. The’ 13 substitutions are 
thus of 5 different types, or say the arrangements to which they 
belong, viz., 
cebad , ceabd , edeab , deabc, eabed , 
edacb , edabc , 
caebd , daebc , 
edbac , debac , 
daecb , deacb . 
are of 5 different types. The question to determine for any value 
of n, the number of the different types, is, it would appear, a diffi- 
cult one, and I do not at present enter upon it. 
