Sec. 3.2 PERMUTATIONS AND COMBINATIONS 65 



from among four. In general, the symbol P„, r stands for the num- 

 ber of r-letter permutations which can be formed from n letters. 

 Then, for n = 4 and r — 2, P 4( 2 = 4(3) = 12, or, in general, 



(3.22) P n , r = n(n - l)(n - 2) • • • (n - r + 1). 



It is possible and useful to express P„, ,- in terms of factorials. To 

 do so, we deliberately create the factorials from formula 3.22 by 

 multiplying and dividing by {n — r) (n — r — 1) ... (2) (1) to make 

 the numerator into n\ and the denominator into (n — r) !. The final 

 result is 



n! 



(3.23) Pn, r = 



(n — ?') ! 



From the definitions of permutations and combinations it follows 

 that every set of r letters can be formed into but one combination, 

 using all r of the letters; whereas, r letters can be formed into r! 

 permutations. Hence, it is concluded that there are r! times as many 

 permutations of n marks taken r at a time as there are different 

 combinations of n letters taken r at a time. Therefore, if the symbol 

 C„, r is adopted to indicate the number of possible combinations of 

 ft letters taken in groups of r letters each, the formula for that num- 

 ber is whichever of 



ft(ft — l)(ft — 2) • • • (ft — r -f- 1) 

 (3.24) C n , r = — r- or 



\O.Zo) wi, r ~ 



we wish to employ. 



r!(ft — r)\ 



Problem 3.21. In how many different orders can 4 cars be parked among 6 

 consecutive parking places along a curb? 



It should be clear in this situation that the order in which the cars 

 are parked makes a difference because the different orders are dis- 

 tinguishable, and would be considered as different by a policeman 

 checking parking. Therefore, this is a problem in numbers of permu- 

 tations and can be worked by either formula 3.22 or 3.23. By 

 formula 3.23 



6! 6-5-4-3-2-1 

 P 6 , 4 = = = 360. 



(6-4)! 2-1 



