50 REY. J. P. KIRKMAN ON THE 
irreversible, it is necessary that m be even, for the lines 7 
must form corresponding pairs. 
We have (Art. 6), 4v-—3 edges k not meeting the base, 
on which to distribute our m points 7, an odd number of 
edges. Now, our subject is such that if any plane per- 
pendicular to the base is drawn to bisect it, and to contain 
no edge k, the configuration on one side of that plane is a 
repetition of that on the other. There is therefore one 
edge k’ bisected by that plane, and for every point 7 planted 
on one half of k we must plant another 7 on the other 
half; and every operation on one side that plane must be 
repeated in the proper place on the other side, if our result 
is to be doubly irreversible. 
We have, then, to consider in how many ways 4m points 
z can be laid on 4(4a-3+4+1)=2e-1 edges k. The number 
of operations by the preceding article is 
m 2a—2|1 
(5 +1) of 
Pen a 
which are to be repeated in direct order on both sides the 
bisecting plane. 
These results are all different, because no sequence can 
be read round the base to begin at two different points 
on one side of our bisecting plane. And no more results 
could be obtained by varying the position of that plane, 
since every possible disposition of $m lines 7 has already 
been made on half the figure. 
If the I°(4v+1,2x) be the entire number of doubly 
irreversible (4v+1)-edra having 2z triangles, and I’I?(4x 
+m-+1,22) be the number of doubly irreversibles thus 
constructible, we have a 
nm 5 +1) 
PP (4x +-m+1,27)=P(S2+1,2e)-2. sap 
Qx—2|1 
for the portion thus obtainable of the whole number 
T’(4e-+m+1,2x2) of doubly irreversible (4¢+m-+1)-edra 
