Richard组合数学第5版-第5章课后习题答案（英⽂版）Math475Text:Brualdi,Introductory Combinatorics5th Ed. Prof:Paul TerwilligerSelected solutions for Chapter51.For an integer k and a real number n,we shown k=n?1k?1+n?1k.First assume k≤?1.Then each side equals0.Next assume k=0.Then each side equals 1.Next assume k≥1.RecallP(n,k)=n(n?1)(n?2)···(n?k+1).We haven k=P(n,k)k!=nP(n?1,k?1)k!.n?1 k?1=P(n?1,k?1)(k?1)!=kP(n?1,k?1)k!.n?1k(n?k)P(n?1,k?1)k!.The result follows.2.Pascal’s triangle begins111121133114641151010511615201561172135352171182856705628811936841261268436911104512021025221012045101···13.Let Z denote the set of integers.For nonnegative n∈Z de?ne F(n)=k∈Zn?kk.The sum is well de?ned since?nitely many summands are nonzero.We have F(0)=1and F(1)=1.We show F(n)=F(n?1)+F(n?2)for n≥2.Let n be /doc/6215673729.htmling Pascal’s formula and a change of variables k=h+1,F(n)=k∈Zn?kk=k∈Zn?k?1k?1=k∈Zn?k?1k+h∈Zn?h?2h=F(n?1)+F(n?2).Thus F(n)is the n th Fibonacci number.4.We have(x+y)5=x5+5x4y+10x3y2+10x2y3+5xy4+y5and(x+y)6=x6+6x5y+15x4y2+20x3y3+15x2y4+6xy5+y6.5.We have(2x?y)7=7k=07k27?k(?1)k x7?k y k.6.The coe?cient of x5y13is35(?2)13 185.The coe?cient of x8y9is0since8+9=18./doc/6215673729.htmling the binomial theorem,3n=(1+2)n=nk=0nSimilarly,for any real number r,(1+r)n=nk=0nkr k./doc/6215673729.htmling the binomial theorem,2n=(3?1)n=nk=0(?1)knk3n?k.29.We haven k =0(?1)k nk 10k =(?1)n n k =0(?1)n ?k n k 10k =(?1)n (10?1)n =(?1)n 9n .The sum is 9n for n even and ?9n for n odd.10.Given integers 1≤k ≤n we showk n k =n n ?1k ?1.Let S denote the set of ordered pairs (x,y )such that x is a k -subset of {1,2,...,n }and yis an element of x .We compute |S |in two ways.(i)To obtain an element (x,y )of S there are n k choices for x ,and for each x there are k choices for y .Therefore |S |=k n k .(ii)Toobtain an element (x,y )of S there are n choices for y ,and for each y there are n ?1k ?1 choices for x .Therefore |S |=n n ?1k ?1.The result follows.11.Given integers n ≥3and 1≤k ≤n .We shown k ? n ?3k = n ?1k ?1 + n ?2k ?1 + n ?3k ?1.Let S denote the set of k -subsets of {1,2,...,n }.Let S 1consist of the elements in S thatcontain 1.Let S 2consist of the elements in S that contain 2but not 1.Let S 3consist of the elements in S that contain 3but not 1or 2.Let S 4consist of the elements in S that do|S |= n k ,|S 1|= n ?1k ?1 ,|S 2|= n ?2k ?1 ,|S 3|= n ?3k ?1 ,|S 4|= n ?3k .The result follows.12.We evaluate the sumnk =0(?1)k nk 2.First assume that n =2m +1is odd.Then for 0≤k ≤m the k -summand and the (n ?k )-summand are opposite.Therefore the sum equals 0.Next assume that n =2m is even.Toevaluate the sum in this case we compute in two ways the the coe?cient of x n in (1?x 2)n .(i)By the binomial theorem this coe?cient is (?1)m 2m m .(ii)Observe (1?x 2)=(1+x )(1?x ).We have(1+x )n =n k =0n k x k,(1?x )n =n k =0nk (?1)k x k .3By these comments the coe?cient of x n in(1?x2)n isn k=0nn?k(?1)knk=nk=0(?1)knk2.2=(?1)m2mm.13.We show that the given sum is equal ton+3k .The above binomial coe?cient is in row n+3of Pascal’s /doc/6215673729.htmling Pascal’s formula, write the above binomial coe?cient as a sum of two binomial coe?ents in row n+2of Pascal’s triangle.Write each of these as a sum of two binomial coe?ents in row n+1of Pascal’s triangle.Write each of these as a sum of two binomial coe?ents in row n of Pascal’s triangle.The resulting sum isn k+3nk?1+3nk?2+nk?3.14.Given a real number r and integer k such that r=k.We showr k=rr?kr?1k.First assume that k≤?1.Then each side is0.Next assume that k=0.Then each side is 1.Next assume that k≥1.ObserverP(r?1,k?1)k!,andr?1k=P(r?1,k)k!=(r?k)P(r?1,k?1)k!.The result follows.15.For a variable x consider(1?x)n=nk=0nk(?1)k x k.4Take the derivative with respect to x and obtain n(1x)n1=nk=0nk(?1)k kx k?1.Now set x=1to get(?1)k k.The result follows.16.For a variable x consider(1+x)n=nk=0nkx k.Integrate with respect to x and obtain(1+x)n+1 n+1=nk=0nkx k+1k+1+Cfor a constant C.Set x=0to?nd C=1/(n+1).Thus (1+x)n+1?1n+1=nk=0nkx k+1k+1.Now set x=1to get2n+1?1 n+1=k+1.17.Routine.18.For a variable x consider(x?1)n=nk=0nk(?1)n?k x k.Integrate with respect to x and obtain(x?1)n+1 n+1=nk=0nk(?1)n?kx k+1k+1+Cfor a constant C.Set x=0to?nd C=(?1)n+1/(n+1).Thus (x?1)n+1?(?1)n+1n+1=nk=0nk(?1)n?kx k+1k+1Now set x =1to get(?1)n n +1=n k =0n k(?1)n ?k 1k +1.Therefore1n +1=n k =0 n k (?1)k 1k +1 .19.One readily checks2 m 2 + m 1=m (m ?1)+m =m 2.Therefore n k =1k 2=nk =0k 2=2nk =0 k 2 +n k =0k1=2 n +13 +n +12 =(n +1)n (2n +1)6.20.One readily checksm 3=6 m 3 +6 m 2 + m1.Thereforen k =1k3=n=6nk =0 k3+6n k =0 k2 +n k =0k1 =6 n +14 +6 n +13 +n +12 =(n +1)2n 24= n +12 2.621.Given a real number r and an integer k .We showrk=(?1)kr +k ?1k .First assume that k <0.Then each side is zero.Next assume that k ≥0.Observe r k =(r )(r 1)···(r k +1)k !=(?1)kr (r +1)···(r +k ?1)k !=(?1)kr +k ?1k.22.Given a real number r and integers k,m .We showr m m k = r k r ?km ?k.First assume that mObserver m m k =r (r ?1)···(r ?m +1)m !m !k !(m ?k )!=r (r ?1)···(r ?k +1)k !(r ?k )(r ?k ?1)···(r ?m +1)(m ?k )!= r k r ?k m ?k .23.(a) 2410.(b) 94 156.(c) 949363.(d)94156949363.24.The number of walks of length 45is equal to the number of words of length 45involving10x ’s,15y ’s,and 20z ’s.This number is45!10!×15!×20!.725.Given integers m 1,m 2,n ≥0.Shown k =0m 1k m 2n ?k = m 1+m 2n .Let A denote a set with cardinality m 1+m 2.Partition A into subsets A 1,A 2with cardinalitiesm 1and m 2respectively.Let S denote the set of n -subsets of A .We compute |S |in two ways.(i)By construction|S |= m 1+m 2n .(ii)For 0≤k ≤n let the set S k consist of the elements in S whose intersection with A 1has cardinality k .The sets {S k }n k =0partition S ,so |S |= nk =0|S k |.For 0≤k ≤n we now compute |S k |.To do this we construct an element x ∈S k via the following 2-stage procedure: stage to do #choices 1pick x ∩A 1 m 1k2The number |S k |is the product of the entries in the right-most column above,which comes to m 1k m 2n ?k .By these comments |S |=n k =0m 1k m 2n ?k .The result follows.26.For an integer n ≥1shown k =1 n k n k ?1 =12 2n +2n +1 ? 2n n .Using Problem 25,n k =1 n k nk ?1 =n k =0n k n k ?1 =n k =0n k nn +1?k =2n n +1 =12 2n n ?1 +12 2n n +1.8It remains to show12 2nn ?1 +12 2n n +1 =12 2n +2n +1 ? 2n n.This holds since2n n ?1 +2 2n n + 2n n +1 = 2n +1n +2n +1n +1= 2n +2n +1.27.Given an integer n ≥1.We shown (n +1)2n ?2=nk =1Let S denote the set of 3-tuples (s,x,y )such that s is a nonempty subset of {1,2,...,n }and x,y are elements (not necessarily distinct)in s .We compute |S |in two ways.(i)Call an element (s,x,y )of S degenerate whenever x =y .Partition S into subsets S +,S ?with S +(resp.S ?)consisting of the degenerate (resp.nondegenerate)elements of S .So |S |=|S +|+|S ?|.We compute |S +|.To obtain an element (s,x,x )of S +there are n choices for x ,and given x there are 2n ?1choices for s .Therefore |S +|=n 2n ?1.We compute |S ?|.To obtain an element (s,x,y )of S ?there are n choices for x,and given x there are n ?1choices for y ,and given x,y there are 2n ?2choices for s .Therefore |S ?|=n (n ?1)2n ?2.By these comments|S |=n 2n ?1+n (n ?1)2n ?2=n (n +1)2n ?2.(ii)For 1≤k ≤n let S k denote the set of elements (s,x,y )in S such that |s |=k .Thesets {S k }nk =1give a partition of S ,so |S |= n k =1|S k |.For 1≤k ≤n we compute |S k |.To obtain an element (s,x,y )of S k there are n k choices for s ,and given s there are k 2ways to choose the pair x,y .Therefore |S k |=k 2 nk .By these comments|S |=n k =1k 2 n k .The result follows.28.Given an integer n ≥1.We shown k =1k n k 2=n 2n ?1n ?1 .Let S denote the set of ordered pairs (s,x )such that s is a subset of {±1,±2,...,±n }andx is a positive element of s .We compute |S |in two ways.(i)To obtain an element (s,x )of S There are n choices for x ,and given x there are 2n ?1n ?1 choices for s .Therefore|S |=n 2n ?1n ?1.9(ii)For1≤k≤n let S k denote the set of elements(s,x)in S such that s contains exactlyk positive elements.The sets{S k}nk=1partition S,so|S|=nk=1|S k|.For1≤k≤nwe compute|S k|.To obtain an element(s,x)of S k there are nkways to pick the positiveelements of s and nn?kways to pick the negative elements of s.Given s there are kways to pick x.Therefore|S k|=k nk2.By these comments |S|=nk=1knk2.The result follows.29.The given sum is equal tom2+m2+m3n .To see this,compute the coe?cient of x n in each side of(1+x)m1(1+x)m2(1+x)m3=(1+x)m1+m2+m3.In this computation use the binomial theorem.30,31,32.We refer to the proof of Theorem5.3.3in the text.Let A denote an antichain such that|A|=nn/2.For0≤k≤n letαk denote the number of elements in A that have size k.Sonk=0αk=|A|=nn/2.As shown in the proof of Theorem5.3.3,≤1,with equality if and only if each maximal chain contains an element of A.By the above commentsnk=0αknn/2nknk≤0,with equality if and only if each maximal chain contains an element of A.The above sum is nonpositive but each summand is nonnegative.Therefore each summand is zero and the sum is zero.Consequently(a)each maximal chain contains an element of A;(b)for0≤k≤n eitherαk is zero or its coe?cient is zero.We now consider two cases.10Case:n is even.We show that for0≤k≤n,αk=0if k=n/2.Observe that for0≤k≤n, if k=n/2then the coe?cient ofαk isnonzero,soαk=0.Case:n is odd.We show that for0≤k≤n,eitherαk=0if k=(n?1)/2orαk=0 if k=(n+1)/2.Observe that for0≤k≤n,if k=(n±1)/2then the coe?cient ofαk is nonzero,soαk=0.We now show thatαk=0for k=(n?1)/2or k=(n+1)/2. To do this,we assume thatαk=0for both k=(n±1)/2and get a contradiction.By assumption A contains an element x of size(n+1)/2and an element y of size(n?1)/2. De? ne s=|x∩y|.Choose x,y such that s is maximal.By construction0≤s≤(n?1)/2. Suppose s=(n?1)/2.Then y=x∩y?x,contradicting the fact that x,y are incomparable. So s≤(n?3)/2.Let y denote a subset of x that contains x∩y and has size(n?1)/2. Let x denote a subset of y ∪y that contains y and has size(n+1)/2.By construction |x ∩y|=s+1.Observe y is not in A since x,y are comparable.Also x is not in A by the maximality of s.By construction x covers y so they are together contained in a maximal chain.This chain does not contain an element of A,for a contradiction.33.De?ne a poset(X,≤)as follows.The set X consists of the subsets of{1,2,...,n}. For x,y∈X de?ne x≤y whenever x?y.Forn=3,4,5we display a symmetric chain decomposition of this poset.We use the inductive procedure from the text.For n=3,,1,12,1232,233,13.For n=4,,1,12,123,12344,14,1242,23,23424,For n=5,,1,12,123,1234,123455,15,125,12354,14,124,124545,1452,23,234,234525,23524,2453,13,134,134535,13534,345.1134.For 0≤k ≤ n/2 there are exactlyn kn k ?1symmetric chains of length n ?2k +1.35.Let S denote the set of 10jokes.Each night the talk show host picks a subset of S for his repertoire.It is required that these subsets form an antichain.By Corollary 5.3.2each antichain has size at most 105 ,which is equal to 252.Therefore the talk show host can continue for 252nights./doc/6215673729.htmlpute the coe?cient of x n in either side of(1+x )m 1(1+x )m 2=(1+x )m 1+m 2,In this computation use the binomial theorem.37.In the multinomial theorem (Theorem 5.4.1)set x i =1for 1≤i ≤t .38.(x 1+x 2+x 3)4is equal tox 41+x 42+x 43+4(x 31x 2+x 31x 3+x 1x 32+x 32x 3+x 1x 33+x 2x 33)+6(x 21x 22+x 21x 23+x 22x 23)+12(x 21x 2x 3+x 1x 22x 3+x 1x 2x 23).39.The coe?cient is10!3!×1!×4!×0!×2!which comes to 12600.40.The coe?cient is9!3!×3!×1!×2!41.One routinely obtains the multinomial theorem (Theorem 5.4.1)with t =3.42.Given an integer t ≥2and positive integers n 1,n 2,...,n t .De?ne n = ti =1n i .We shownn 1n 2···n t=t k =1n ?1n 1···n k ?1n k ?1n k +1···n t.Consider the multiset{n 1·x 1,n 2·x 2,...,n t ·x t }.Let P denote the set of permutations of this multiset.We compute |P |in two ways.(i)We saw earlier that |P |=n !n 1!×n 2!×···×n t != n n 1n 2···n t.12(ii)For1≤k≤t let P k denote the set of elements in P that have?rst coordinate x k.Thesets{P k}tk=1partition P,so|P|=tk=1|P k|.For1≤k≤t we compute|P k|.Observe that|P k|is the number of permutations of the multiset{n1·x1,...,n k?1·x k?1,(n k?1)·x k,n k+1·x k+1,...,n t·x t}. Therefore|P k|=n?1n1···n k?1n k?1n k+1···n t.By these comments|P|=tn1···n k?1n k?1n k+1···n t.The result follows.43.Given an integer n≥1.Show by induction on n that1 (1?z)n =∞k=0n+k?1kz k,|z|<1.The base case n=1is assumed to hold.We show that the above identity holds with n replaced by n+1,provided that it holds for n.Thus we show1(1?z)n+1=∞=0n+z ,|z|<1.Observe1(1?z)n+1=1(1?z)n11?z=∞k=0n+k?1kz k∞h=0z h=0c zwherec =n?1+n1+n+12+···+n+ ?1=n+.The result follows.1344.(Problem statement contains typo)The given sum is equal to (?3)n .Observe (?3)n =(?1?1?1)n=n 1+n 2+n 3=nnn 1n 2n 3(?1)n 1+n 2+n 3=n 1+n 2+n 3=nnn 1+n 2+n 3=nnn 1n 2n 3(?1)n 2.45.(Problem statement contains typo)The given sum is equal to (?4)n .Observe (?4)n =(?1?1?1?1)n=n 1+n 2+n 3+n 4=nnn 1n 2n 3n 4(?1)n 1+n 2+n 3+n 4=n 1+n 2+n 3+n 4=nnn 1n 2n 3n 4(?1)n 1?n 2+n 3?n 4.Also0=(1?1+1?1)n= n 1+n 2+n 3+n 4=nnn 1n 2n 3n 4(?1)n 2+n 4.46.Observe√30=5=5∞ k =01/2k z k.For n =0,1,2,...the n th approximation to √30isa n =5n k =0 1/2k 5?k.We have14n a n051 5.52 5.4753 5.47754 5.47718755 5.477231256 5.4772246887 5.4772257198 5.4772255519 5.477225579 47.Observe101/3=21081/3=2(1+z)1/3z=1/4,=2∞k=01/3kz k.For n=0,1,2,...the n th approximation to101/3isnk=01/3k4?k.We haven a n021 2.1666666672 2.1527777783 2.1547067904 2.1543852885 2.1544442306 2.1544327697 2.1544350898 2.1544346059 2.15443470848.We show that a poset with mn+1elements has a chain of size m+1or an antichain of size n+1.Our strategy is to assume the result is false,and get a contradiction.By assumption each chain has size at most m and each antichain has size at most n.Let r denote the size of the longest chain.So r≤m.By Theorem5.6.1the elements of the posetcan be partitioned into r antichains{A i}ri=1.We have|A i|≤n for1≤i≤r.Thereforemn+1=ri=1|A i|≤rn≤mn, 15for a contradiction.Therefore,the poset has a chain of size m+1or an antichain of size n+1.49.We are given a sequence of mn+1real numbers,denoted{a i}mni=0.Let X denote the setof ordered pairs{(i,a i)|0≤i≤mn}.Observe|X|=mn+1.De?ne a partial order≤on X as follows:for distinct x=(i,a i)and y=(j,a j)in X,declare xof{a i}mni=0,and the antichains correspond to the(strictly)decreasing subsequences of{a i}mni=0sequence{a i}mni=0has a(weakly)increasing subsequence of size m+1or a(strictly)decreasingsubsequence of size n+1.50.(i)Here is a chain of size four:1,2,4,8.Here is a partition of X into four antichains:8,124,6,9,102,3,5,7,111Therefore four is both the largest size of a chain,and the smallest number of antichains that partition X. (ii)Here is an antichain of size six:7,8,9,10,11,12.Here is a partition of X into six chains:1,2,4,83,6,1295,10711Therefore six is both the largest size of an antichain,and the smallest number of chains that partition X.51.There exists a chain x116。
