ii8 Algebra. 



It is further evident that if we form all possible arrangements in 

 each group we thereby obtain the total number of arrangements 

 of the 71 letters taken ?- at a time. 



The total number of arrangements then equals the number of 

 arrangements in each group multiplied by the number of groups. 

 Hence, representing the number of groups of n things taken r at 

 a time by G(") and remembering that the number of arrange- 

 ments in each group equals the number of arrangements of r 

 things taken all at a time, that is ' ?', and further remembering 

 that the total number of arrangements equals 



n(n—\)(n—2) . . (n — r-\-\), 



we have 



11 



hence GC") = — , 



5. The form of this result shows that the number of groups of 

 u things taken r at a time is the same as the number taken « — r 

 at a time. This is also evident in another way, for every time we 

 select f things from ii things we leave out ?z — r things ; hence there 

 must be as many ways oi leaving out 7i — r things as of selecthig r 

 things, but of course there are as many ways of selectiug n — r 

 things as there are oi leaving out n — r things. 



6. In all that precedes, it was supposed that the given things 

 were all different and that in forming the arrangements or groups 

 none of the given things were repeated. Now we will consider 

 arrangements and groups in which the things maj^ be repeated 

 and those in which the given things are not all alike. 



7. Problem. To find the Number of Arrangements op 

 71 Things taken r at a Time, Repetitions being Allowed. 



Suppose first we wivSh the number of arrangements, including 

 repetitions, of .the four letters a, b, c, d taken one at a time. 

 Evidently there are four arrangements, viz : a, b, c, d. 



Next suppose we wish the arrangements, including repetitions, 

 of the four letters a, b, c, d taken two at a time. 



