DERIVED FROM THE REGULAR POLYTOPES. 43 



The equations + x-, + x i = 3 give 24 forms (2 1) [10], i.e. 

 24 jP 4 of edge import. 



The equations 2 + ^ = 4 give 10 forms (2 1 1 0), i. e. 10 CO 

 of body import. 



So, we find 24 CO and 24 C, i.e. 48 polyhedra, and therefore 

 1 (24 X I 4 + 24 X 6) = 240 faces. 



So the result is (90, 288, 240, 48) in accordance with the law 

 of Euler. 



Example [32 110], direct method. 



The number of vertices is 2 4 . 5! divided by 2!, i. e. 1920:2 

 = 960. 



The edges split up into three groups (32), (21), (10). Through 

 the pattern vertex pass: one edge (32), two edges (21) — on 

 account of the two digits 1 — and four edges (10) — on account 

 of the two digits 1 and of the faculty to make the last digit to 

 correspond either to -f- x b or to — œ b . 



So there are in toto 



480 edges (32), 960 edges (21), 1920 edges (10), 

 i. e. 3360 edges. 



The faces split up into six groups, viz. the triangles (211) and 

 (110), the squares (32) (10), (21) (10) and [10] and the hexagon 

 (321). 



In the pattern vertex concur : 



one triangle (211), 



two triangles (110), on account of + œ b , 



four squares (32) (10), on account of 'the two digits 1 and of + x h 



(21) (10), , 

 two „ [10], „ 



two hexagons (321), „ „ „ „ 1 ). 



So we find: 



/ 3 triangles 10 squares 2 hexagons 



960 ^— _ -\ + _ 



= 960 triangles -\~ 2400 squares — |- 320 hexagons, 



i. e. 3680 faces. 



The limiting bodies split up into the seven groups: 



(3211) = tT, (321) (10) = P 6 , (32) (110) = P 3 , (2110) = CO 

 (32) [10] = (21) [10] = P 4 , [110] = CO. 



5J JJ 55 J5 )J ?3 35 3? 33 3 

 33 33 33 33 33 3 



z ) In the case [10] the difference between +#5 and — x§ has no effect, on account 

 of the square brackets. 



