104 
ON THE PARTITIONS OF A POLYGON. 
[915 
915] 
But we must exclude the distributions where there is no side in the one interval r — 4 
and no side in the other interval between the two diagonals; the number of these [ n w j- 
is that for the case of the coalescence of the two diagonals into a single diagonal, 
viz. it is 
_ i— 3 
= 1 ’ But 
and thus the number required is diago] 
r — 3.1— 2. r — 1 r — 3 
6 1 ' 
18. Three diagonals, A.—There must be on each side of the three diagonals, 
that is, in two of the six intervals formed by the diagonals, two sides; there remain 
r — 4 sides to be distributed between the same six intervals, and the number of ways ^ 
in which this can be done is 
each 
r — 3 . r — 2. r — 1. r. r + 1 „ 
= . v — D 
120 , 
m wl 
But we must exclude distributions which would permit the coalescence of the first 
and second, or of the second and third, or of all three of the diagonals. For the 
coalescence of the first and second diagonals (the third diagonal not coalescing), the ^ 
term to be subtracted is re 
_ _ , tl form i 
r — 3.1— 2 .o— 1 r — 6 
6 r ; 
and the same number for the coalescence of the second and third diagonals (the 
first diagonal not coalescing); that is, the last-mentioned number is to be multiplied which 
by 2; and for the coalescence of all three diagonals the number to be subtracted is ^ 
_ r — 3 each 
1 v — 8 
we have thus the foregoing value which 
v — 3.7— 2 .r — l.r.r+1 _ r — 3. r — 2. r — 1 r — 3 
120 2 - 6 + 1 ’ 
where it will be observed that we have the binomial coefficients 1, 2, 1 with the which 
signs 2 
Three diagonals, B.—There must be outside each of the three diagonals, that is, t 
in three of the six intervals formed by the diagonals, two sides; and there remain which 
v — 6 sides to be distributed between the six intervals; the number of ways in which have 
this can be done is diagor 
r — 5 . r — 4 .r — 3. v — 2.0' — 1 as to 
120 ’ diagor 
and there is here no coalescence of diagonals, so that this is the number required. diago 
19. Four diagonals, A.—There must be on each side of the four diagonals, that ^ us 
is, in two of the eight intervals formed by the diagonals, two sides; there remain eac ^