THEORY OF THE PARTITIONS OF NUMBERS. 53 



must be rejected if they involve X or p. as coefficients, because by hypothesis we are 

 only concerned with I X's and m //.'s, and these have already been accounted for. 



Observe now that the columns which contain a symmetrical selected pair of X's 

 only contain /t's which are in the same rows as these X's, and therefore the deletion of 

 these columns cannot delete p.'s appertaining to any rows except those occupied by 

 the selected pair of X's. Observe further that the column which contains an 

 unsymmetrical X, say in the p ih row, contains a /A in the 2np+I a> row, and that 

 therefore the disappearance of a p. in the 2n p+l th row follows from the deletion of 

 a column containing an unsymmetrical X in the p th row. 



Hence of the 2n I m rows in question 



l + m %i 2j rows contain '2nlml, a elements, 

 and thence 



2n 2l 2m+2i+2j rows contain '2iilm2, a. elements. 



Accordingly if s is the sum of all the a elements except those which appear as 

 coefficients of the selected X's and /A'S the co- factor of 



n\ in-i\ 9 ,- 2i in-l 



l-2i 

 contains l + m2i2j factors of type 



and 2n 2l 2m + 2i + 2j factors of type 



(*--,,), 

 or ofnlm + i+j squared factors of type 



(s-oi v -.)*, 



since these factors occur in equal pairs. 



Hence the co-factor is 



n(*-a,,)ll(.s-a,.-a,,.) 2 , 



wherein the quantities a u , l + m2i2j in number, which appear in the first product, 

 and the quantities ., 2n2l-2m + 2i+2j in number, which appear in the second 

 product, are all different. 



Also (s <* a K ,) 2 is effectively equal to 



s 2 2 ( + a,,.) s + 2a v a. u , 



since squares of the a's may be rejected. 



Hence, by the reasoning employed in the first solution we may put the quantities 

 a equal to unity, regard s p as equal to p ! symbolically, and say that the coefficient of 



in the product 



