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XXIII. — On a Problem in Combinations. By The Rev. Philip Kelland, M.A., 

 Professor of Mathematics in the University of Edinburgh. 



(Read 3d December 1855.) 



Several years ago, when discussing the question of the distribution of the stars, 

 a problem occurred to Professor Forbes, which, simple as it is, appears to have 

 escaped notice prior to that time. Having been consulted as to its solution, I 

 communicated my results to Professor Forbes, who has inserted one of them in 

 his paper printed in the Philosophical Magazine for 1850, vol. xxxvii., p. 425. 

 But for the very ingenious application which Professor Forbes has there made of 

 it, the problem might probably not be worth recurring to. As it is, I have thought 

 it would not be altogether uninteresting to give the complete solution. 



The Problem is as follows : — There are n dice, each of which has p faces, p 

 being not less than n ; it is required to find the number of arrangements which 

 can be formed with them, 1°, So that no two show the same face ; 2°, That no three 

 show the same face ; 3°, That no four do so, and so on. 



1 . The number of arrangements in which no two show the same face is easily 

 seen to be the same as the number of permutations of the p faces, taken n to- 

 gether; and is therefore p (p-l) (p- 2) . . . . (p—n + 1). 



2. Remove the dice A and B, and cover the face 1 of the remainder. The 

 number of arrangements which these can now form, omitting the covered face, 

 and no two showing the same face, will be — 



(p-1) 0-2) .... (p-l-n-2 + 1) 

 = 0-1) 0-2) .... (p-n + 2). 



Place with each of these the dice A and B, showing face 1, and you have the 

 arrangements in which the dice A and B, and these alone, show face 1. The same 

 applies to each of the other faces. Consequently there are p (p—l) • • • • {p—n + 2) 

 rrangements in which the dice A and B, and these alone, show the same face. 

 The same is true of every other pair of dice. Hence the number of arrangements 

 in which two, and two only, show the same face, is — 



n 12 p (^ -1 ) • ' • ' (P~ n + 2 )- 



3. Remove the dice A, B, C, D, and cover the faces 1 and 2 of the others. 

 The number of arrangements which can now be formed in which no two show the 

 same face is — 



O- 2 ) OP" 3 ) • • • • (p-2-n=3 + l) 



= 0-2) 0-3) (p-n + 3). 



VOL. XXI. PART III. 5 E 



