494 
Proceedings of Royal Society of Edinhurgh. 
the series of pairs of first suffixes in every row and the series of 
pairs of second suffixes in every column being 
12, 13, 14, 15, 23, 24, 25, 34, 35, 45 ; 
that is to say, the combinations arranged in ascending order, of 
the numbers 1, 2, 3, 4, 5, taken two at a time. On the first side 
of the identity are 10 products, and as both factors of each product 
contain 10 terms, the result of the multiplification would be to 
produce 1000 terms of the form ” 
the whole expansion in fact being 
q=b S=5 w=5 
2 2 2 I • 
<?=2 s=2 n^2 
Pcq r<s m^n 
On the right hand side are 100 terms of the form 
and if a proof of the identity were wanted, we should only have to 
show that each of the 100 terms of the latter kind gives rise to a 
particular 10 terms of the former kind. This, too, it is interesting 
to note, Cauchy himself could have done. For example, the last 
of the 100 terms, 
l™44«*66l 
^41®^41 ^'42®42 + • • • + ^45®45 - ^41®’51 4 <^42^32 d " • • • + <^45®'55 
^51^^41 d" %2®^42 d- . . • + %5«45 ^51*^51 d* ^52^52 + • • • + *^55“55 ’ 
^41 
«42 
(X43 
«44 
«45 
«51 
^52 
%3 
«54 
%5 
«41 «42 HS ®'44 ®^45 
“■51 ^52 ®’53 ®54 ®55 ’ 
«42 
“•41 “42 
+ 
^41 ^43 
“41 “43 
4- + 
«44 
“44 “45 
^52 
“51 “52 
^51 ®53 
“51 “53 
^54 ^55 
“54 “55 
which is nothing more than Cauchy’s formula (62) 
fi.v y v.i fi.i j , 
when we put /x=10 = i/, and p=2. Instead of 1000 terms on the 
left-hand side and 100 on the right, we should clearly have for the 
general theorem terms on the left and terms on the right, P 
be it remembered being the combinatorial 
