PROBLEM IN COMBINATIONS. 361 
_n(n—1) ,n(m—l)....-. (n= 3)» 
Be dae p—n+1 ¥ (p—n+1) (p— =e 
n(n—1)....(m—5) 1 , & 
hp 1 = —n4+a) oer c. 
d p-nt1du ped ta p—n H(i ies . (n—8) p—ntl 
dal? Tz) me, ApEn ly 
nm (n—1)....(n—5) a sais ee 
(p—n+1) Gz nm+2)1.2 ; 
Qn-2p nN. n—8) 2n—2p—2 
=n(n—1)z ry Se + &e. 
when reset 
a 
? 2p—n—2 a p—at1idu & a n a 
Hence z aia a Ta) => ae (¢ u) 5 
2p—n— = n—2 0 u —2p+2n Au 
or of {2 2p+2 e+ (eH wee pt qa} 
= 42° S46) tS an (wl) Pw; 
whence eee a + {p- n+1+ (4n— 6)al Se — —n(n—1) u=0. 
7. To find the number of arrangements in which no quadruplications occur, 
let us write the result of Art. 5 under the form C;, being the total number of 
arrangements of m 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,_ 7; is the 
number of arrangements of those dice in which no triplications occur; and, 
4 consequently, the number in which A, B, C, and those only, show face 1. Hence 
p oa 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— 
tesa Ps 
8. Remove the dice A, B, C, D, E, F, and cover faces 1 and 2 of the others ; 
, —_ is the number of arrangements in which A, B, C show face 1, and D, E, F 
6.5.4.3.2.1 , p (p—l) ons 6 
(1 2.3) 1.2 asi 
