CH. x] kikkman's school-girls problem 223 



which x girls walking in rows of a abreast can be arranged so 

 that every possible combination of b of them may walk abreast 

 once and only once is as yet unsolved. Suppose that this number 

 of ways is y. It is obvious that, if all the x girls are to walk out 

 each day in rows of a abreast, then x must be an exact multiple 

 of a and the number of rows formed each day is x/a. If such 

 an arrangement can be made for z days, then we have a solution 

 of the problem to arrange x girls to walk out in rows of a 

 abreast for z days so that they all go out each day and so that 

 every possible combination of b girls may walk together once 

 and only once. For instance, if x = 2n, a = 2, b = 2, we have 

 y = n(2n-l)j2, z = 2n-l. If * = 15, a = 3, 6 = 2, we have 

 y = 35 ; and these 35 rows can be divided into 7 sets, each of 

 which contains all the symbols ; hence z = 7. 



