ii4 The Rev. Thos. P. Kirkman on 



We have no /-pie prime in our belt. The least 2-ple is 

 Fig 6 ; the least 3-ple is Fig 7. Observe that by 2-ple we 

 mean always zoneless 2-ple 



If we had used a prime having 12 marginal triangles, 

 our belts, of x— five primes each, would be numbered not by 

 half dozens, but probably by hundreds, all as much alike in 

 features as are the above 18 equivalents of one selected belt. 



We shall speak in Art. 8 of the fi permutations of their 

 order in the belt of the x—^ primes, which are quite 

 different from the above M permutations of the parts of all 

 the m integral k-partitions of the mere number x. The 

 numbers k and x may differ in any way or degree. Here 

 x = k=$ is convenient. 



Each of the 18 belts will add 39 of its 41 not crossed 

 (Art. 1) summits to the 23 of H (Art. 3), and 19 faces, of 

 which 14 are marginal triangles, to the 15 faces of H. 



At this point it is requisite that we clearly state the 

 complete problem that we intend to solve. It is this — to 

 enumerate the number of partitioned R-gons, that all alike 

 fulfil the conditions following : — 



1st. That they be all reducible to the same irreducible 

 H, or to an irreducible identical with H, in the names of 

 the polygons that compose H, two of which shall be exactly 

 the 3-gon and the 5-gon that have for edges the dark 

 diagonals in H. 



2nd. That they each contain one of the 18 equivalent 

 belts above described. 



3rd. That they all contain the same tape, Fig. 8. 



Observe, that no partitioned R-gon can contain more 

 than one selected belt or tape, still less two irreducibles H. 



Our problem resembles a famous old one : In how many 

 ways can you put all of/ things into n fixed places, leaving 

 any z>o of the n places vacant? In that, when e things are 

 wanted to put in places chosen, any e of the / things, if 

 they are all units, will serve as well as any other e of 



