ii6 Algebra. 



Now multiply these equations together, member b}^ member, 

 and we get 



Aj Ao A,3 . . . A„=2Ai 3A2 . . . wA,,_i 



= 1x2x3 . . . w Ai A., . . . A„_i. 

 By cancelling common factors we get 



A;,= i X2X3 . . . n. 

 The product of the integer numbers from 71 down to i or from 

 I up to n is often represented by |^^ ox 7i\ , and is read factorial ?i, 

 or 71 admiration. 



With this notation we may write 



A,=k. 



3. Problem. To find the Number of Arrangements of 

 ;^ Things taken r at a Time. 



ivCt us first take a particular case, say the number of arrange- 

 ments of five things, say the five letters a, d, c, d, e, taken three 

 at a time. . Suppose the arrangements all made and we select 

 those which begin with a and put them by themselves in one 

 class, then those which begin with d and put them by themselves 

 in another class, and so on. We then divide the whole number 

 of arrangements into five classes, and it is evident that the num- 

 ber in any one class is just the same as in any other class. 

 Consider those which begin with a. Then every arrangement in 

 this class contains besides a two of the four letters d, c, d, e, and 

 since a is fixed and the other letters arranged in every possible 

 way, therefore the number of ^/tese arrangements must equal 

 the number of arrangements of the four letters d, c, d, c taken 

 two at a time. 



Iti ge7ieral, if we have 7t things, say the letters a, b, c,d,e,f, . . 

 to be taken r at a time, we may select all those arrangements 

 which begin with a and put them by themselves in one class, then 

 those which begin with <^ and put them by themselves in another 

 class, and so on. We thus divide the whole number of arrange- 

 ments into 71 classes, and it is evident that the number of 

 arrangements in any one class is just the same as the number of 

 arrangements in any other class. 



Consider those which begin with a. 



