Dr T. Muir on Simple Alternants . 
1887 .] 
437 
the number of dovetailings would be increased by 2 + 3 + 3 + 2 + 1, 
i.e. 11; so that the coefficient in this instance would be 46. 
In finding the coefficient of 2,a 10 b 9 c 5 d 5 the only new point of 
difference would be an additional pair, 5, 6. This would cause 
1 + 1 + 1 + 1 additional dovetailings, and would consequently make 
the coefficient 46 + 4, i.e. 50. 
The method may sometimes be used to obtain more general 
results. Thus — 
Third Example , — Required the coefficient of %a s ~ 7 b 2 c 2 d 2 in the 
expansion of a( 0, 1, 4, s), where of course s- 7 >2, that is, s>9. 
0 
1 2 3 
4 5 s- 1 
A 
B 
c 
Duads: 0, 
5; 
1, 4; 2, 3. 
Triads: 0, 
1, 8; 
0, 2, 7; 
0, 3, 6. 
Humber of dovetailings; 3 + 2 + 1, i.e. 6, which is the coefficient 
required, and which, be it observed, is independent of s. 
In this way have been obtained several important theorems, to 
which I shall now pass. The first is — 
The expansion of | cd > b l S‘d ?J+s | + 1 a°b l c 2 d 3 | or a( 0 1 3 3 + s) con- 
sists of all the single symmetric functions ivhicli have no index 
greater than s, and the sum of whose indices is s+ 1, the coefficient 
of each function being less by 1 than the number of different letters 
appearing in any of its terms. (i.) 
Tor example, to find the expansion of | a°b l c^d 7 | -f- 1 a% l c 2 d 3 | or 
a(0 13 7) we subtract each index of the divisor from the corre- 
sponding index of the dividend, and thus are led to the first of the 
symmetric functions in the quotient, viz. Then writing 
downthe succeeding symmetric functions %a% 2 ^a%c, %a 2 b 2 c i % a 2 bcd , 
and prefixing to each a coefficient less by 1 than the number of 
letters appearing in any term of the function, we have 
a( 0 13 7) = + %a% 2 + 2 ta%c + T%a 2 b 2 c + 3 ta 2 bcd. 
The proof is as follows: — The only functions which can occur in 
the expansion must be of the form 'Sa a bP, %a a lAcy, fa a b^cyd 8 ; conse- 
quently all that we have to do is to determine the coefficients of 
