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XXIIT.—On a Problem in Combinations. By The Rev. Putte Ketianp, 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 Forpes, 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 Forses 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 m dice, each of which has p faces, p 
being not less than 2; 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—1) (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) (p—2).. . .(p—1—n—2+1) 
=p} ie= 2). -(p—n+2). 
Place with each of these the dice A and B, showing face 1, and you ene 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—1)....(p—n+2) 
_ trangements 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— 
joe np (p— 1)... .(p—n+2). 
3. Remove the dice A, % 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— 
—2) (p—3).... (p—2—n—441) 
=(p—2) (p—3).... (p—n+t3). 
VOL. XXI. PART III. SE 
