PROBLEM IN COMBINATIONS. 361 



T . n du n (n-1) n Q-l) (n-B) 



llien d^-p-n + 1 + (p-n + 1) (p-n + 2) 



n(n-l) (n-5) J_ 2 & 



+ (»-k + 1). . . .(p-n + B) 1.2 



(p — n + 1). . . . (_p — 7i + 3) 



/ p-n+ldl 



dec 



+ 



d / p-n+ldu\ , -v P-n ,«.... (n-3) P-n+1 

 I CO 3 — I =« (W— 1) ^ + -. pfv # 



j> — n+2 



w(n-l).. . . (w-5 ) x &( , 



(p — w+1) (p — w + 2) 1.2 



. -, 2n-2p U . . . . (n — 3) 2n-2p-2 „ 



v ' p — n + L 



when a? = —s' 



2p-n— 2 f — 2»+2re— 2 <Z 2 W , -, ,. — 2p + 2» ^M"| 



or z [s __, + b _ n + i), — j 



. n — 6d 2 U ,. „ N n — 4 C? W , 1N w— 2 



* d"P~ ( 6 ^ d^ + n<Jl >* W; 



whence (#— 4<» 2 ) =— 5 + ■< p—n + 1 + (4 n— 6 )<# I ^ n (»— 1) w=0. 



7. To find the number of arrangements in which no quadruplications occur, 

 let us write the result of Art. 5 under the form (J being the total number of 

 arrangements of n dice, each having p faces, in which no triplications occur. 

 Remove the dice A, B, C, and cover the face 1 of the others. Then C^_ x is the 

 number of arrangements of those dice in which no triplications occur ; and, 

 consequently, the number in which A, B, C, and those only, show face 1. Hence 

 p C n _ 1 is the number of arrangements in which a triplication is found on the 

 first three dice, and on those alone. Consequently the number of arrangements 

 in which one triplet only occurs is — 



n (n-1) (w-2) p «-3 

 1.2.3 p S-i' 



8. Remove the dice A, B, C, D, E, F, and cover faces 1 and 2 of the others ; 

 Cpl 2 is the number of arrangements in which A, B, C show face 1, and D, E, F 

 face 2. 



Hence 6.5.4.3.2.1 «-« 



1.2.3.1.2.3 p ~ 2 



is the number of arrangements in which A, B, C, D, E, F have faces 1 and 2 



tripled; consequently 



6.5.4.3.2.1 , p Q-l) p «-6 

 (1 2.3/ 1.2 p- 2 



