Méthodes mathématiquesdel’informatiqueII Spring2007 Notes on Combinatorics Claude Auderset University of Fribourg 1. Ordered choices Various combinatorial problems can be viewed as involving the choice of k objects from a set of n objects. The number of ways of making such a choice depends on whether or not a given object may be chosen more than once, and whether or not the order of choice is important. Ordered choices with repetitions 1.1 Theorem. The number of lists (a ,...,a ) of k (non necessarily distinct) ele- 1 k ments of a set of n elements is nk (k,n N). This number is to be taken 1 if k =0 ∈ (including the case where also n=0). Proof. The first element a may be chosenin n ways,the secondone a in n 1 2 ways,andsoon, k times. So there aren n n n=nk possibilities. When k =0, the only list of length 0 is the empty list·. · ··· (cid:3) Such lists are also called k-tuples. They can be interpreted as words of length k on an alphabet of n letters, ⋄ orderedsamplesofk ballsdrawnfromanurnofn ballswith replacement. ⋄ Forinstance,thereare265 =11881376fiveletterwordsusingtheRomanalphabet, and 28 =256 eight bit bytes. Let A be a set of n elements and B = b ,...,b a set of k elements. There 1 k { } is a one-to-one correspondence (bijection) between the set AB of maps f :B A → and the set Ak of lists (a ,...,a ) of k elements of A. To each map f : B A 1 k → correspondsthelist(f(b ),...,f(b )) Ak and,conversely,alist(a ,...,a ) Ak 1 k 1 k ∈ ∈ defines a map f AB by f(b ) = a (i = 1,...,k). Via this correspondence, the i i ∈ theorem above translates into 1.2 Theorem. Given finite sets A and B, AB = A|B|, | | where X denotes the cardinality ((cid:12)num(cid:12)ber of elements) of a finite set X. (cid:3) | | (cid:12) (cid:12) Classical interpretations: there are nk colourings of k distinct objects with a palette of n colours (each object ⋄ being painted any one of the n colours), distributions of k distinct objects into n boxes (not considering the order ⋄ of objects in the boxes). 2 Section 1. Ordered choices The power set P(X) of a set X consists of all subsets of X (including the empty set ). There is a one-to-one correspondence between P(X) and 0,1 X. To each ∅ { } subset A of X corresponds its indicator function (also known as its characteristic function) χ :X 0,1 A −→{ } 1 if x A, x ∈ 7−→(0 if x / A. ∈ Conversely, to each function f : X 0,1 corresponds the subset A = x X → { } { ∈ | f(x)=1 . Using the previous result, we get P(X) = 0,1 X =2|X|. } | | { } 1.3 Theorem. A set of n elements has 2n subsets (n (cid:12)N). (cid:12) (cid:3) ∈(cid:12) (cid:12) Ordered choices without repetitions 1.4 Theorem. The number of lists (a ,...,a ) of k distinct elements of a set of n 1 k elements is given by the falling factorial k [n] =n(n 1) (n k+1)= (n i+1) (k,n N). k − ··· − − ∈ k times iY=1 This number is ta|ken to be 1{zif k =0 (i}nclusively when n=0 too), and 0 if k >n. Proof. Suppose that 0 < k n. The first element a may be chosen in n 1 ≤ ways, the second one a in only n 1 ways since it must be different from a , and 2 1 − soon. Thei-thchoice(i=1,...,k)isfromthen (i 1)remainingelementsonly. − − Whenk =0,thereisexactlyonelistoflengthk,namely the emptylist. When k >n, there is no list of k distinct elements among n elements. (cid:3) Recall that a map f : B A is injective when f(b) = f(b′) b = b′ or, → ⇒ equivalently, b=b′ f(b)=f(b′) for all b,b′ B. With this terminology, 6 ⇒ 6 ∈ 1.5 Theorem. Let A and B be finite sets with A =n and B = k. The number of injective maps from B into A is [n] . | | | | (cid:3) k Classical interpretations: there are [n] k orderedsamples ofk balls drawnfroman urnofn balls (without replace- ⋄ ment), ordered hands of k cards from a deck of n cards, ⋄ distributions of k distinct objects into n boxes subject to the exclusion ⋄ principle, i.e., every box containing at most one object. In the case A = n = k = B , a map f : B A is injective if and only if it | | | | → is surjective, i.e., for every a A there is some b B such that f(b) = a. So, ∈ ∈ in this case, the three notions ‘injective’, ‘surjective’ and ‘bijective’ (= injective + surjective) coincide. A permutation is a bijective map from a set into itself. 1.6 Theorem. The number of permutations of a set of cardinality n N is given ∈ by the factorial n n!=[n] =1 2 n= i. n · ··· ntimes Yi=1 This number is to be taken 1 if n=0. (cid:3) | {z } For instance, the letters of the word ‘STOP’ can be arranged in 4!=24 ways. Section 2.Unordered choices 3 Let us mention without proof the following way of computing the factorial of large numbers. 1.7 Stirling’s approximation n n n! √2πn (n N,n>0) ≈ e ∈ (cid:16) (cid:17) with a relative error <1/11n. 1.8 Theorem. The number of distributions of k distinct objects into n boxes, considering the order of the objects in each box, is given by the raising factorial k [n]k =n (n+1) (n+k 1)= (n+i 1) (k,n N). · ··· − − ∈ k times iY=1 This number is t|aken to be 1{zif k =0. } Proof. Thefirstobjectmaybeputinoneoftheboxesinnways. Thesecond object may be put either in the same box as the firstobject with 2 possible positions, left or ⋄ right of the first object, or in one of the n 1 other boxes. ⋄ − So, there are 2+(n 1)=n+1 ways of distributing the second object into one of − the n boxes. Letuslookingeneralathowmanywaysthereareofdistributingthei-thobject into one of the boxes. At that time, i 1 objects have already been distributed, say m objects into the j-th box (j =1−,...,n, n m =i 1). The i-th object j j=1 j − may be put in m +1 ways in the j-th box j P either by placing it just before one of the m objects in the box, j ⋄ or by appending it after the m objects. j ⋄ All together,the number of waysofputting the i-thobjectin one of the n boxes is n n (m +1)= m +n=(i 1)+n=n+i 1. j j   − − j=1 j=1 X X   The number of distributions of k objects into n boxes is the product of these numbers where i runs from 1 to k. (cid:3) 2. Unordered choices Unordered choices without repetitions 2.1 Theorem. The number of subsets of cardinality k of a set of cardinality n is given by the binomial coefficient n [n] n! = k = (k,n N). k k! (n k)!k! ∈ (cid:18) (cid:19) − This number is taken to be 1 if k =0, and 0 if k >n. Proof. Everysubset a ,...,a ofk distinctelementsofasetAofcardinal- 1 k { } ity n gives rise to k! lists (a ,...,a ) by permuting the k elements a ,...,a . As i1 ik 1 k thereare[n] listsofkdistinctelementsofA,thenumberofsubsetsofcardinalityk k of A is [n] /k!. (cid:3) k 4 Section 2.Unordered choices A subsetofcardinalityk ofa setofcardinalityn is alsocalleda k-combination of n objects. Classical models: there are n k (unordered) samples of k balls(cid:0)dr(cid:1)awn from an urn of n balls (without ⋄ replacement), (unordered) hands of k cards from a deck of n cards, ⋄ (ordered) lists (a ,...,a ) of k elements of an ordered set of cardinality 1 k ⋄ n which are (strictly) increasing, i.e., such that a

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