CH. IX] UNICURSAL PROBLEMS 179 



which are either blank or marked with 1, 2, 3... dots. An 

 ordinary set contains 28 dominoes marked 6-6, 6-5, 6-4, 6-3, 

 6-2, 6-1, 6-0, 5-5, 5-4, 5-3, 5-2, 5-1, 5-0, 4-4, 4-3, 4-2,' 4-1,' 

 4-0, 3-3, 3-2, 3-1, 3-0, 2-2, 2-1, 2-0, 1-1, 1-0, and 0-0. 

 Dominoes are used in various games in most, if not all, of 

 which the pieces are played so as to make a line such that 

 consecutive squares of adjacent dominoes are marked alike. 

 Thus if 6-3 is on the table the only dominoes which can be 

 placed next to the 6 end are 6-6, 6-5, 6-4, 6-2, 6-1, or 6-0. 

 Similarly the dominoes 3-5, 3-4, 3-3, 3-2, 3-1, or 3-0, can 

 be placed next to the 3 end. Assuming that the doubles 

 are played in due course, it is easy to see that such a set of 

 dominoes will form a closed circuit*. We want to determine 

 the number of ways in which such a line or circuit can be formed. 

 Let us begin by considering the case of a set of 15 dominoes 

 marked up to double-four. Of these 15 pieces, 5 are doubles. 

 The remaining 10 dominoes may be represented by the sides 

 and diagonals of a regular pentagon 01, 02, &c. The intersec- 

 tions of the diagonals do not enter into the representation, 



and accordingly are to be neglected. Omitting these from our 

 consideration, the figure formed by the sides and diagonals of 

 the pentagon has five even nodes, and therefore is unicursal. 

 Any unicursal route (ex. gr. 0-1, 1-3, 3-0, 0-2, 2-3, 3-4, 4-1, 

 1-2, 2-4, 4-0) gives one way of arranging these 10 dominoes. 

 Suppose there are a such routes. In any such route we may 

 put each of the five doubles in any one of two positions (ex. gr. 



* Hence if we remove one domino, say 5-4, we know that the line formed by 

 the rest of the dominoes must end on one side in a 5 and on the other in a i. 



12—2 



