915] 
ON THE PARTITIONS OP A POLYGON. 
Ill 
123, 222, and so on. Of course the number of the partitions 15, 24, 33, is equal 
to the whole number of the 2-partitions of the 8-gon, that is, =20; the number of 
the partitions 114, 123, 222, is equal to the whole number of the 3-partitions of 
the 8-gon, that is, it is =120; and so in other cases. It is easy to derive in 
order one from the other the numbers of the partitions of each several kind of the 
polygons of the several weights 2, 3, 4, 5, 6, &c.; and I write down the accompanying 
Table (No. 26), facing page 112, the process for the construction being as follows: 
27. The first column (2 parts) is at once obtained. For a polygon of an odd 
number of sides, for instance the 9-gon (weight = 7), imagining the summits numbered 
in order 1, 2, ..., 9, we divide this into a triangle and octagon, or obtain the 
partitions 16, by drawing the diagonals 13, 24,..., 81, 92: viz. the number is =9. 
In the Table this is written, 16 = 9; and so in other cases. Similarly we divide it 
into a quadrangle and heptagon, or obtain the partitions 25, by drawing the diagonals 
14, 25, ..., 82, 93: viz. the number is again =9; and we divide it into a pentagon 
and a hexagon, or obtain the partitions 34, by drawing the diagonals 15, 26, ..., 83, 94: 
viz. the number is = 9, and here 
9 + 9 + 9 = 27, 
the whole number of 2-partitions of the 9-gon. For a polygon of an even number 
of sides, for instance the 10-gon (weight = 8), the process is a similar one, the only 
difference being that for the division into two hexagons, (that is, for the partitions 
44), each partition is thus obtained twice, or the number of such partitions is 
110, =5; the numbers for the partitions 17, 26, 35, 44, thus are 10, 10, 10, 5; and 
we have 
10+10 + 10 + 5 = 35, 
the whole number of the 2-partitions of the 10-gon. 
28. To obtain the second column (3 parts)—suppose, for instance, the 3-partitions 
of the 9-gon; these are 115, 124, 133, 223. We obtain the number of the partitions 
115 from the terms 
16 = 9 and 25 = 9 
of the first column: viz. in 16, changing the 6 into 15, that is, dividing the polygon 
of weight 6 into two parts of weights 1 and 5 respectively: this can be done in 
eight ways (see, higher up, 15 = 8 in the first column); and we thus obtain the 
number of partitions 
9 x 8 = 72; 
and again, in 25, changing the 2 into 11, that is, dividing the polygon of weight 
2 into two parts each of weight 1: this can be done in two ways (see, higher up, 
11 = 2 in the first column); and we thus obtain the number of partitions 
9x 2 = 18; 
we should thus have, for the number of partitions 115, the sum 
72 + 18 = 90, 
only, as it is easy to see, each partition is obtained twice, and the number of the