ON A PROBLEM OF AERANGEMENTS. 
248 
[669 
all the substitutions which move every letter. Thus when n = 5, we obtain the 44 sub 
stitutions for the letters abode, viz. these are 
{abode), &c., 24 symbols obtained by permuting in every way the four letters 
b, c, d, <?; 
(ab) (ode), &c., 20 symbols corresponding to the 10 partitions ab, ode, and for each 
of them 2 arrangements such as ode, ced. 
And then if we reject those symbols which contain in any ( ) two consecutive letters, 
we have the substitutions which give the arrangements 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, when 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) (bee), 
(adeeb), (be) (ado). 
Here in the first column, performing on the symbol (acbed) the substitution (abode), 
we obtain (bdeae), = (aebdc), the second symbol; and so again and again operating 
with (abode), we obtain the remaining 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 difficult one, and I do not at present 
enter upon it.