915] 
ON THE PARTITIONS OF A POLYGON. 
113 
32. To fix the ideas, I assume p = 4, and thus consider the problem of the division 
of the (2w + 2)-gon into quadrangles. Writing 
w — 1 = a+ b + c, 
we take at pleasure any one side of the (2w 4- 2)-gon, making this the first side of 
a quadrangle, the second, third, and fourth sides being diagonals of the polygon such 
that outside the second side we have a polygon of weight a, outside the third side 
a polygon of weight b, and outside the fourth side a polygon of weight c. Any one 
or more of the numbers a, b, c may be = 0; they cannot be each of them = 0 
except in the case w = 1. The meaning is that the corresponding side of the 
quadrangle, instead of being a diagonal, is a side of the (2w + 2)-gon, viz. there is 
no polygon outside such side. Suppose, in general, that P w is the number of ways 
in which a polygon of weight w can be divided in quadrangles, and let each of the 
polygons of weights a, b, c respectively, be divided into quadrangles: the number of 
ways in which this can be done is P a P b P c ; and it is to be noticed that, if for 
instance a= 0, then the number is — P b P c , viz. the formula remains true if only we 
assume P„=l. The number of partitions thus obtained is £P a P b P c , where the 
summation extends to all the partitions of w — 1 into the parts a, b, c (zeros 
admissible and the order of the parts being attended to). And we thus obtain all 
the partitions of the (2w + 2)-gon into w parts; for first, the partitions so obtained 
are all distinct from each other, and next every partition of the (2w -f 2)-gon into w 
parts is a partition in which the selected side of the (2w + 2)-gon is a side of 
some one of the quadrangles. That is, we have 
P w — %P a P b P c (a, b, c as above); 
and it hence appears that, considering the generating function 
f= 1 + P 1 x + P%x 2 + P s x 3 + ..., 
we have 
/= 1 + æ f 3 - 
The reasoning is precisely the same if, instead of a division into quadrangles, we 
have a division into p-gons; the only difference is that instead of the three parts 
a, b, c, we have the p — 1 parts a, b, c, and the equation for / thus is 
/= 1 + x P~ l - 
33. Writing for a moment 
f — U + xfP- 1 , 
and expanding by Lagrange’s theorem, we have 
1.2... w K 
1.2 ... w 
viz. after the differentiation, writing u = 1, we have 
[(p-l)w] w ~ l 
[ w ] W -i 
where it will be recollected that, for the number of sides of the polygon, we have 
r = (p— 2)w + 2. 
In the case of the partition into triangles, p = 3, and we have the before-mentioned 
value 
C. XIII. 
[2w] w_1 + [w\ w ~\ 
w = r — 2. 
15