212 Proceedings. 



3lR5 = 3"3 + 2-6 = 21, 



which is the third coefficient in 



(l + l)*^"*"^"^ = I + 7 + 21 +. . . . 



Again, 



■000005 000014 0001 13 111002 



IIIIIO 000023 000221 =6Q6 



are the 7 6-partitions of 5, whose permutation symbols are, 

 indices being read as multipliers, 



icv'b + 2a'^bc + ■^cC'b'''c, 

 giving the product, 



6lR5=2-6-t- 2-30 + 3-60 = 252, 

 which is the 6th coefficient in 



(i + i)^"*"*^"^ = I + 10 + 45 + 120 + 210 + 252 +. . . 



" These examples, even if they w^ere a thousand, are no 

 rigorous proof of the above theorem. We require a non- 

 tentative demonstration. Such a one I have, but it pleases 

 me neither by its brevity nor by its elegance. I content 

 myself, for the present in this brief abstract, with begging 

 the aid of mathematicians in finding a neater proof before 

 trial, for every x and k, or, if possible, a disproof. 



"Thistheorem,sent by me withoutdemonstration, appeared 

 in the Educational Times in August last. My study of these 

 permutations of the partitions of the number R has given 

 me the mastery of a far more difficult problem, the k- 

 partitions of the R-gon. I shall soon have the honour to 

 present to this Society the correct answer to the following 

 question : — 



"In how many different ways, symmetric or not, can an 

 R-gon be partitioned hy k- i diagonals, none crossing 

 another, into k faces, viz., into a^ triangular, di 4-gonal, 

 as 5-gonal, &c., &c., faces, so that a diagonal or diagonals 

 shall occupy every angle of the R-gon, except the vertices 

 of m marginal triangles, w<or = ^3? 



