30.3 PERMUTATIONS AND COMBINATIONS
We note that (30.27) may be written in a more general form ifSis not simply
divided intoAandA ̄but, rather, intoanyset of mutually exclusive eventsAithat
exhaustS. Using the total probability law (30.24), we may then write
Pr(B)=
∑
i
Pr(Ai)Pr(B|Ai),
so that Bayes’ theorem takes the form
Pr(A|B)=
Pr(A)Pr(B|A)
∑
iPr(Ai)Pr(B|Ai)
, (30.28)
where the eventAneed not coincide with any of theAi.
As a final point, we comment that sometimes we are concerned only with the
relativeprobabilities of two eventsAandC(say), given the occurrence of some
other eventB. 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 combinations
In equation (30.5) we defined the probability of an eventAin a sample spaceSas
Pr(A)=
nA
nS
,
wherenAis the number of outcomes belonging to eventAandnSis the total
number of possible outcomes. It is therefore necessary to be able to count the
number of possible outcomes in various common situations.
30.3.1 Permutations
Let us first consider a set ofnobjects that are all different. We may ask in
how many ways thesenobjects may be arranged, i.e. how manypermutationsof
these objects exist. This is straightforward to deduce, as follows: the object in the
first position may be chosen inndifferent ways, that in the second position in
n−1 ways, and so on until the final object is positioned. The number of possible
arrangements is therefore
n(n−1)(n−2)···(1) =n! (30.30)
Generalising (30.30) slightly, let us suppose we choose onlyk(<n) objects
fromn. The number of possible permutations of thesekobjects selected fromn
is given by
n(n−1)(n−2)···(n−k+1)
︸ ︷︷ ︸
kfactors
=
n!
(n−k)!
≡nPk. (30.31)