392 
PROFESSOR CAYLEY OK" THE PROBLEM OF 
Case 30. e=D=F=x, a=c=y. By reciprocation of 29. 
No. = 2X(X- S)(x-2)y(y- 1)B. 
On account of the symmetry we must divide by 2. 
Case 31. c=e=T)=x, a=B—y. 
X=(Y —2)x(X—2)(x—3)'Fy, % ' = Fa(X-2)(a- 3)Y(y-2), 
g=^-3)(X-2)F{(Y-2)^+Y(y-2)} 
=2x(x- 3)(X- 2)(yY -y- Y)F. 
Case 32. F=B=a=#, D=0=y. By reciprocation of 31. 
No. =2X(X- 3 )(a- 2){yY-y- Y)c. 
Case 33. B=F=y, «=e=D=a\ By reciprocation of 34. 
No.=2(a’— 1 )(aX— a— X)Y(Y— l)c. 
Case 34. c=e=y, B=D=«=a\ (In the Table, erroneously, B=D=e= 
x ={X-2)y{X-l)(y-l)Fx, %' = F/yX (3 / — 1 ) (X — 1 ) (a — 2 ) , 
g=y(y-l)(X-l){(X-2) a r+X(X-2)}F 
= 2(X — 1) (a.’X — a’ — X)y(y — 1 )F. 
Case 35. a=l)=y, c=e= B=a\ 
X = X(a’— 2) Y (x — l)Fy, ^=FxY{x-l){X-2)y, 
g=yY(x-l){X(x-2) + (X-2)x}F 
=2{x—l) (xX — x —X)yY F. 
Case 36. a = D=y, B=F=e=x. By reciprocation of 35. 
No. = 2(X- 1 ){Xx-x-X)yYc. 
Case 37. a—e=Y)=x, c—B=y. By reciprocation of 38. 
No. = 2 (a- — 1 ) [xyXY — xy(X+ Y)— XY(a;+y) + 2#y+2XY} F. 
Case 38. B— D=a—x, F =e=y. 
%— (X— 2)c(X— l)y(Y— 2)a?, ^=Y(y-2)Xc(X-l)(x-2), 
g=(X-ljc{ a y(X-2)(Y-2)+XY(j?-2Xy-2)} 
= 2 (X — 1 ) { xyXY — xy{X + Y) — XY (x -\-y)-\- 2*xy -j- 2X Y } c. 
Case 39. a=c=e=B=x. 
Functional process ; the curve is assumed to be the aggregate of two 
a=o=e=B=x- \-x’. Forming the enumeration 
