Mathematical Methods for Physics and Engineering - Matematica.NET

30.3 PERMUTATIONS AND COMBINATIONSWe note that (30.27) may be written in a more general form if S is not simplydivided into A and Ā but, rather, into any set of mutually exclusive events A i thatexhaust S. Using the total probability law (30.24), we may then writePr(B) = ∑ iPr(A i )Pr(B|A i ),so that Bayes’ theorem takes the formPr(A|B) =Pr(A)Pr(B|A)∑i Pr(A i)Pr(B|A i ) , (30.28)where the event A need not coincide with any of the A i .As a final point, we comment that sometimes we are concerned only with therelative probabilities of two events A and C (say), given the occurrence of someother event B. From (30.26) we then obtain a different form of Bayes’ theorem,Pr(A|B)Pr(C|B) = Pr(A)Pr(B|A)Pr(C)Pr(B|C) , (30.29)which does not contain Pr(B) at all.30.3 Permutations and combinationsIn equation (30.5) we defined the probability of an event A in a sample space S asPr(A) = n A,n Swhere n A is the number of outcomes belonging to event A and n S is the totalnumber of possible outcomes. It is therefore necessary to be able to count thenumber of possible outcomes in various common situations.30.3.1 PermutationsLet us first consider a set of n objects that are all different. We may ask inhow many ways these n objects may be arranged, i.e. how many permutations ofthese objects exist. This is straightforward to deduce, as follows: the object in thefirst position may be chosen in n different ways, that in the second position inn − 1 ways, and so on until the final object is positioned. The number of possiblearrangements is thereforen(n − 1)(n − 2) ···(1) = n! (30.30)Generalising (30.30) slightly, let us suppose we choose only k (< n) objectsfrom n. The number of possible permutations of these k objects selected from nis given byn!n(n − 1)(n − 2) ···(n − k +1) =} {{ } (n − k)! ≡ n P k . (30.31)k factors1133