434 Proceedings of Royal Society of Edinburgh, [july 18 , 
The quotient 
1 
a 3 
a 6 
1 
a 
a 2 
1 
b 3 
b 6 
1 
b 
b 2 
1 
c 3 
c 6 
1 
c 
c 2 
i.e. | «°5 3 c 6 1 4 - 1 a% l c 2 j , 
or, say, A(0, 3, 6) A(0, 1, 2) , 
or, still more shortly, <z(0, 3, 6) , 
is evidently a homogeneous symmetric function of degree 
3 + 6 -(1 + 2), i.e. 6, and when expressed as a sum of single 
symmetric functions will be of the form 
%cdb 2 + x%odbc + y^,a 3 b s + z%a% 2 c + u%a 2 b 2 c 2 . 
The problem is to determine x, y, z, u. 
Professor Johnson’s method is to use repeatedly his above- 
mentioned reduction-theorem. Tims 
a(0,3,6) = H 42 + a5cH 2>1 + a 2 b 2 c 2 H 00 , 
= Xa^b 2 + %a^bc + % a 3 b s + ^ a% 2 c + %a 2 b 2 c 2 \ 
+ abc(1a 2 b + %abc) > 
+ a 2 b 2 c 2 . ) 
= Z2a 4 5 2 + %a^bc + %a 2 b ? ’ + 2 %a% 2 c + 3 %a 2 b 2 c 2 . 
My method contrasts with this in that it determines the co- 
efficients x } y, z, u separately. Besides, therefore, being of interest 
as throwing a side-light on Professor Johnson’s method, it may be 
found useful when only one or a very few coefficients are wanted, 
and it has certainly been the means of arriving at several more or 
less noteworthy results. 
The basis of it is the expansion of the alternants in terms of 
alternants of lower orders. 
Alternants of Third Order. 
First Example . — Required the coefficient of a% 2 c in the ex- 
pansion of a( 0, 3, 6). 
Solution . — Write the integers from 0 to 6, and separate them 
into three groups A, B, C, by putting a bar before 0, 3 and 6 ; thus 
o l i 
3 4 5 
A 
B 
6 . 
C 
