1887 .] 
Dr T. Muir on Simple Alternants. 
435 
In the group A delete those which are greater than the index of c 
in %a% 2 c. Find in how many ways a number from group A, with 
a number from group B, will give the sum 4, 4 being 1 more than 
the sum of the last two indices in %ci d b 2 c. This number of times, 
2, is the coefficient required. 
The solution is here put at greater length than need be, in order 
that when taken with the corresponding solution for the case of 
alternants of the fourth order, the generality of the method may be 
apparent. It may be enunciated quite shortly as a theorem, viz.: — 
The coefficient of %a x b v c z in the expansion of \ afh p c q \ -7- 1 a% l c 2 1 is 
the number of ways in 'winch by taking a number from 
0, 1 , 2, . . . , z-1, z 
and a number from 
P,P+ 1, • • • , - 1 
the sum y + z + 1 may be obtained. 
Second Example. — Required the coefficient of %a%^c 2 in the ex- 
pansion of a( 0, 3, 8). 
Solution. — The two groups here are 0, 1, 2 and 3, 4, 5, 6, 7; 
and the sum 6 can be made up from them in three ways, viz. 
0 + 6, 1 + 5, 2 + 4. The coefficient therefore is 3. 
Alternants of Fourth Order. 
First Example. — Required the coefficient of %a^b 2 c 2 d in the 
expansion of a( 0, 2, 4, 9). 
Solution. — Write the integers from 0 to 9, and separate them 
into four groups A, B, C, D, by putting a bar before 0, 2, 4 and 9 ; 
thus 
0 1 
2 3 
4 5 6 7 8 
A 
B 
C 
In group A delete, if necessary, those which are greater than the 
index of d in ^a*b 2 c 2 d. Take the integers from 0 to the index of 
d inclusive, and unite each with such a number as will make the 
sum 4, that is to say, 1 more than the sum of the last two indices 
of '%a it b 2 c 2 d : this gives 
0,4; 1,3. 
Take three numbers, one from A, one from B, and one from C, 
