915] 
ON THE PARTITIONS OF A POLYGON. 
103 
6 
4 diagonals, A = 
r — 3.r — 2. r — l.r.r+l.r + 2.r + 3 
5040 
Q r—3.r — 2.r — l.r.r + 1 r — 3.r—2.r — 1 r — 3 
-3 J2Ö + 3 g l-j- 
r — 3.r — 4.r — 5 .r — 6 
5040 
(r 3 + 18r 2 + 65?") ; 
where the calculation is 
?" — 2 . r — 1. r. r + 1 ,r + 2. r + 3 = r 6 + 3r 6 — 5r 4 — 15r®+ 4r 2 + 12r 
— 126. r — 2 . r — 1. r. r + 1 — 126r 4 + 252r 3 + 126r 2 — 252?" 
+ 2520. r - 2. r - 1 + 2520r 2 - 7560r + 5040 
- 5040 - 5040 
r 6 + 3r 5 - 131r 4 + 237r 3 + 2650?" 2 - 7800r 
= r — 4.r — 5.r — 6.r 3 + 18?" 2 + 65?". 
Also 
B = 
r — 5.r — 4.?’ — 3 . r — 2.r — l.r.?" + l r — 5 .r — 4.r — 3.r — 2 .r — 1 
5040 
r — 3.r — 4.r — 5 . r — 6 
5040 
120 
(?" — 1. r — 2. ?" + 7), 
„ ?" — 7.?"—6.r — 5.?" — 4.?" — 3.?" — 2.?"—1 
\J = 
5040 
r — 3.r — 4.r — 5.r — 6 
5040 
where the calculation is 
(r — 1. ?" — 2. r — 7); 
r.r+1 =?- 2 + r 
- 42 - 42 
r 2 + r — 42 
= r — 6 .r + 7. 
17. To explain the formation of these expressions, observe that : 
One diagonal.—There must be on each side of the diagonal, or say in each of 
the two “ intervals ” formed by the diagonal, two sides ; there remain ?" — 4 sides 
which may be distributed at pleasure between the two intervals, and the number of 
ways in which this can be done is 
?" — 3 
Two diagonals.—There must be on each side of the two diagonals, or say in 
two of the four intervals formed by the diagonals, two sides; there remain r — 4 
sides to be distributed between the same four intervals, and the number of ways in 
which this can be done is 
?• — 3.?" — 2.r — 1