242 Proceedings of Royal Society of Edinburgh. [sess. 
where | 0 a /3 y 8 | denotes the determinant 
| A° B“ O' 5 D T E s | . 
The problem now is to determine X, /x, .... to satisfy the re- 
lation. 
To show that any one of the products cannot exist, it will be 
sufficient to show that it contains a term which is contained in no 
other product. 
The term B 5 C 2 D 3 E 12 d 2 cb is contained in the product 
| a?Wc 2 D 4 eE 4 | . | 01238 | , 
and the only other product in which it might seem to appear is 
| a°5 1 c 2 D 2 eE 2 | • | 012510 j, but a litfle investigation will show 
that it does not. It follows, therefore, that v b = 0 . 
Similarly, on account of the terms 
A 4 B C 6 D 4 E 7 e 2 d a , 
B 2 C 7 D 8 E 5 e 2 db, 
BC 7 D 8 E 6 e 2 dh, 
A 4 B C 6 D 5 E 6 e 2 dc, 
which the products 
| a 0 b l c 2 D 4 eE 4 | • | 01247 
| a 9 c 2 D 4 eE 4 | • | 02345 | , 
| b 1 c 2 D 4 eE 4 | • | 01346 | , 
| a« b 1 c 2 D 4 eE 4 | • | 01256 |, 
respectively contain, and which are contained in no other products, 
we have v 4 = v 3 = v 2 = v 1 = 0 . 
Since v 4 = v b = 0, it follows that-/x 9 = /x ]0 = /x n = g l2 = /x 13 = 0 , for 
these terms are the only ones which contain eleventh and twelfth 
powers of the large letters. 
Again, on account of the term^B C 6 D 6 E 10 d 2 c b, which is found 
in | a 0 b 1 c 2 1) 2 eE 2 | • | 01368 | , and in no other product, 
/x 2 = 0. 
The term B C 8 D 3 E 10 e d 2 b is contained in the product 
| a 0 b 1 c 2 B 6 eE 6 | • | 01234 | , and since /x 2 and v 2 , the coefficients 
of the only other products which could contain this term, vanish, 
it follows that p = 0 . The term B 2 C 6 D 5 E 9 d 2 c b is contained in 
