24 G EULEE. 



440. As an example of Euler's method we will give his inves- 

 tigation of the case in which three tickets are drawn. 



There are three events which may happen which may be repre- 

 sented as follows : 



I. a, a-\-l, a-\-2, that is a sequence of three. 



II. a, a + 1; h, that is a sequence of two, the number h 

 being neither a + 2 nor a — 1. 



III. a, h, c, where the numbers a, Jj, c involve no sequence. 



I. The form a, a-\-l, a + 2. The number of such events is 

 n — 2. For the sequence may be (1, 2, 3), or (2, S, 4), or (3, 4, 5), 

 up to (71 —2,n—l, n). 



II. The form a, a + 1, h. In the same way as we have just 

 shewn that the number of sequences of three, like a, a + 1, « + 2, 

 is 71 — 2, it follows that the number of sequences of two, like 

 <2, a + 1, is 7z — 1. Now in general h may be any number between 

 1 and n inclusive, except a—1, a, a + 1, a + 2; that is, h may be 

 any number out of ?? — 4 numbers. But in the case of the first 

 sequence of two, namely 1, 2, and also of the last sequence n — 1, 71, 

 the number of admissible values of J is n — 3. Hence the whole 

 number of events of the form a, « + 1, I, is (w — 1) (71 — 4) + 2, that 

 is 71^ — 5n + Q), that is {n — 2) {ii — 3). 



III. The form a, h, c. Suppose a to be any number, then h 

 and c must be taken out of the numbers from 1 to a — 2 inclusive, 

 or out of the numbers from a + 2 to n inclusive ; and b and c must 

 not be consecutive. Euler investiofates the number of events 

 which can arise. It will however be sufficient for us here to take 

 another method which he has also given. The total number of 

 events is the number of combinations of 7i things taken 3 at a time, 



that is — ^^ — — . The number of events of the third kind 



can be obtained by subtracting from the whole number the num- 

 ber of those of the first and second kind ; it is therefore 



71 (n — 1) Cii — 2) , J,. , ,,. . 



17273 — ~ ^'' ~ ^ ^" ~ ) ~ ^'' " ^- 



