64 ELEMENTARY PROBABILITY Ch. 3 



involve different letters. Two sets of three (or of n) letters are said 

 to be different permutations of letters if they differ in either the order 

 in which the letters are arranged, or if some different letters are in- 

 volved in the two sets. Two groups of n letters are considered to be 

 different combinations of letters only if some different letters are 

 included in the two sets. A reordering of the same letters forms a 

 new permutation but not a new combination. Hence, just one com- 

 bination can be formed from n given letters if all the letters are used 

 at once. This question then arises: How many different permuta- 

 tions can one form from n letters, using all n letters each time? 



The process of constructing a permutation consists in determining 

 a first, a second, . . . , and finally an nth letter. The first letter is 

 chosen from among n, the second from among the remaining (n — 1) 

 letters, the third from among the (n — 2) then remaining unchosen, 

 etc., until finally only one letter is left for the nth choice. These n 

 choices can be made in n(n — 1) (n — 2) (n — 3) ... (2) (1) ways, 

 which is then the number of different permutations possible with n 

 letters if all n of them are used in each permutation. To illustrate, 

 suppose that there are three letters: A, B, and C. The following out- 

 line shows how the choices can be made: 



Therefore the permutations are: ABC, ACB, BAC, BCA, CAB, CBA, 



and there are 3(2) (1) = 6 of them. 



It is convenient to denote the product n(n — 1) (n — 2) . . . (2) (1) 

 by the symbol n !, and to call it n factorial. Hence, if P n , „ is adopted 

 as the symbol for the number of permutations of n marks arranged in 

 sets of n, we have 



(3.21) P Bi „ = n! 



as the formula for computing the number of such permutations. If 

 n = 3, as above, P 3 , 3 = 3! = 3-2-1 = 6, as before. 



More often, it is necessary to make up permutations of marks in 

 which only r of the n marks are used at any one time. For example, 

 we might be choosing a batting order of 9 men from a squad of more 

 than 9 men. To see how the process goes, suppose that it is required 

 to make up all the possible two-letter permutations from the letters 

 A, B, C, and D. There are 4 choices for the first letter and 3 choices 

 for the second; therefore, there are 4(3) = 12 possible choices of two 



