410 
REV. T. P. KIRKMAN ON A CLASS OF POLYEDRA. 
Next to find the hendecaedra, we write down 
1(11, 5) =R(6, 2).IR(6, 2, 3)=0; 
R(ll, 5) — R(6, 2).RR(6, 2, 3) ; 
R(l], 4) = R(7, 2).RR(7, 2, 2) + R®(7, 3).RR^(7, 3, 1); 
1(11, 4) =R(7, 2).IR(7, 2, 2)+P(7, 2).IP(7, 2, 2) + R^(7, 3).IR^(7, 3, 1) ; 
P(11,4) = P(7, 2).PP(7, 2, 2); 
R(ll, 3) = R(8, 2).RR(8, 2, 1) ; 
1(11, 3) =R(8, 2).IR(8, 2, 1) + I(8, 2). 11(8, 2, 1) + I(8, 3). 11(8, 3, 0) ; 
R(ll, 2) = R(9, 2).RR(9, 2, 0) ; 
1(11, 2) =R(9, 2).IR(9, 2, 0) + I(9, 2). 11(9, 2, 0) + P(9, 2).IP(9, 2, 0) ; 
P(ll, 2)=P(9, 2).PP(9, 2, 0). 
That is — 
R(ll, 5) = 1 .m(3, 1, 1) = 1 ; 
R(ll, 4) = l.f«(4, 1, 1) + 1.2 =5; 
1(11, 4) 2, 2)-z7(4, 1, 1)} 
+ l.i{n(7, 2, 2)-iii4, 1, 1)} + 1.1 =11 ; 
P(ll, 4) = l.n(4, 1, 1) = 3 ; 
R(ll, 3) = 2.f/(4, 1, 0) = 4 ; 
1(11,3) =2.i{z7(8,2, 1 )-m(4, 1,0)} 
+ l.u(8, 2, 1) + 1 .m(8, 3, 0) 
= 2.7+1.16+1.8 =38; 
R(ll, 2) = 2.77(5, 1, 0)=2.2 =4; 
1(11, 2) =2.i{'«(9, 2, 0) — n(5, 1, 0)}+2.i«(9, 2, 0) 
+ 2.1{z7(9, 2, 0)-^7(5, 1, 0)}=2. 1+2. 4 + 2. 1 = 12 ; 
P(ll, 2)=2.z7(5, 1, 0) = 4. 
As a verification, it may be worth while to write down these eighty-two 1 1-edra, to 
show the faces in order about the 10-gonal base. 
R(ll, 5) is 3537353636; 
R(1 1, 4) are 5383535453, 4383453635, 3464363636, 4373537345, 3463536437; 
1(11,4) are 4438353635, 6346436363, 5437436363, 3543835354, 3637374354, 
6353653463, 6353644373, 6346354373, 5353736345, 6353644373, 
4437363536; 
P(ll, 4) are 5353653536, 6346363463, 5437354373; 
