组合数学(第5章5.2)
组合数学第五版答案
组合数学第五版答案简介《组合数学第五版答案》是对组合数学第五版的习题答案进行整理和解答的参考资料。
组合数学是一门研究集合之间的组合方式和规律的数学科学。
它广泛应用于计算机科学、统计学、运筹学等领域，在算法设计、图论分析等方面有着重要的应用价值。
本文档包含了《组合数学第五版》中各章节的习题答案，主要内容涵盖了排列组合、图论、生成函数、递推关系、容斥原理等多个重要主题。
通过对这些习题的解答，可以帮助读者更好地理解组合数学的基本概念、方法和应用。
目录•第一章：基本概念和方法•第二章：排列组合•第三章：图论•第四章：生成函数•第五章：递推关系•第六章：容斥原理第一章：基本概念和方法1.习题1：证明排列的总数为n! (阶乘)。
2.习题2：计算组合数C(n, m)的值。
3.习题3：探究组合数的性质并给出证明。
第二章：排列组合1.习题1：计算排列数P(n, m)的值。
2.习题2：解决带有限制条件的排列问题。
第三章：图论1.习题1：证明图论中的握手定理。
2.习题2：解决图的着色问题。
第四章：生成函数1.习题1：利用生成函数求解递推关系。
2.习题2：应用生成函数解决组合数学问题。
第五章：递推关系1.习题1：求解递推关系的通项公式。
2.习题2：应用递推关系解决实际问题。
第六章：容斥原理1.习题1：理解容斥原理的基本思想并给出证明。
2.习题2：应用容斥原理解决计数问题。
结论通过对《组合数学第五版答案》中的习题进行解答，读者可以更好地掌握组合数学的基本概念和方法。
组合数学在计算机科学、统计学、运筹学等领域具有广泛的应用，通过学习和理解组合数学，读者可以提高解决实际问题的能力，并为进一步深入研究相关领域打下坚实的基础。
注：本文档中的习题答案仅供参考，请读者在独立思考和解答问题时加以思考和验证，以深入理解组合数学的核心概念和方法。
组合数学(引论)
组合数学中有二个常用的技巧： 1. 一一对应 2. 奇偶性
1.、一一对应
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1. 一一对应
二个事件之间如计果算存：在一一对应关系，则
可用解易解的来替代第难一解轮的：。50场比赛 (一人轮空)
应用举例 第二轮： 25场比赛 (一人轮空)
决出例冠1军. 共有要10进1行个注反一多选第第第意之场少手三四五：，比场参轮轮轮每要赛比加：：：场淘。赛象1比汰63？棋3场场场赛一淘比比比必 人汰赛赛赛淘也赛汰必，((一 一一须问人 人人进要轮 轮，行空 空))
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3. 幻方
3. 幻方
2）麦哲里克方法 （与德拉鲁布方法类似）
将1置正中央上方，然后按向右上方的方向依次放后 继数； 到顶行后翻到底行，到达最右列后转最左列； 其余情况放正上方2格。
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3. 幻方
3. 幻方
2）麦哲里克方法 （与德拉鲁布方法类似）
将1置正中央上方，然后按向右上方的方向依次放后 继数； 到顶行后翻到底行，到达最右列后转最左列； 其余情况放正上方2格。
第4章 Burnside引理与Polya定理
4.1 群的概念 4.2 置换群 4.3 循环、奇循环与偶循环 4.4 Burnside引理 4.5 Polya定理 4.6 鸽巢原理 4.7 鸽巢原理举例 4.8 鸽巢原理的推广 4.9 Ramsey数
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一、一组、合组数合学数简学介简介
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总统 副总统 财务大臣 秘书
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1
2
2
43
2
1
一种选法 一一对应 一个四位数
组合数学与图论
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第2章 图论基础
什么是图论
图论是研究图结构的 数学分支，用于描述 对象之间的关系。图 由节点和边组成，节 点表示对象，边表示 对象之间的关系。
基本概念
无向图
边没有方向的图
权重图
边带有权重的图
度
节点相连的边数 称为节点的度
91%
有向图
边有方向的图
图的表示方法
01 邻接矩阵
02 邻接表
判断图中的节点是否都是连通的
02 组合数学方法
连通性定理和算法可以用于判断和求解
03
总结
组合数学和图论相互结合，能够解决图的同构、 着色、匹配和连通性等各种问题，通过组合数学 方法的运用，可以更好地探索图论中的难题。
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第四章 组合数学与图论在计 算机科学中的应用
图数据库与图搜索
图数据库是一种专门用于存储和查询图结构数据 的数据库系统。在计算机科学中，图搜索算法如 Dijkstra算法、A*算法等被广泛应用于图数据库 的查询和分析，帮助用户快速准确地获取所需信 息。
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第五章 组合数学与图论在统 计学中的应用
基于图的统计分 析
利用组合数学和图论 的方法进行统计学分 析，如图的频繁模式 挖掘、图数据的聚类 分析等。这些方法能 够帮助研究人员从大 量数据中提取出有用 的信息并进行深入分 析。
网络数据采样与推断
节点采样
通过在网络中随 机选择节点来获
取样本数据
使得相邻节点颜 色不同
图的匹配问题
图的匹配问题是指在 图中找到一些相互不 相邻的边，使得边的 数量最大化。组合数 学的匹配定理和匹配 算法可以用于解决图 的匹配问题。
图的连通性问题
Richard组合数学第5版-第5章课后习题答案（英文版）
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。
Richard组合数学第5版-第5章课后习题答案(英文版)
evaluate the sum in this case we compute in two ways the the coefficient of xn in (1−x2)n. (i)
By the binomial theorem this coefficient is (−1)m
2m m
.
(ii) Observe (1 − x2) = (1 + x)(1 − x).
5
Now set x = 1 to get Therefore
(−1)n n =
n (−1)n−k 1 .
n+1
k
k+1
k=0
1
n
=
n (−1)k 1 .
n+1
k
k+1
k=0
19. One readily checks
m 2
+
m
= m(m − 1) + m = m2.
2
1
Therefore
n
n
k2 =
not contain 1 or 2 or 3. Note that {Si}4i=1 partition S so |S| =
4 i=1
|Si|.
We
have
n
n−1
n−2
n−3
n−3
|S| =
, k
|S1| =
, k−1
|S2| =
, k−1
|S3| =
, k−1
|S4| =
. k
The result follows.
We have
n P (n, k) nP (n − 1, k − 1)
=
组合数学第五章
{}1 ,,n a n b n c n a ⋅⋅⋅、从中取出个字母，要求的个数为偶数，问有多少种取法？
2 a b c d e n a b 、由字母，，，，组成的长为的字中，要求与的个数之和为偶数，问这样的字有多少个？
{}123412123,,,1,2,3,41,2,3,41,2,451,2,3,410n i i i S e e e e a S n e i e i e e e e e i ∞⋅∞⋅∞⋅∞⋅===、设多重集合=，表示集合满足下列条件的组合数，分别求数列的生成函数:
(1)每个()出现奇数次；
(2)每个()出现3的倍数次；
（3）不出现，至多出现次；
（4）出现13或11次，出现或次；
（5）每个()至少出现次。
{}124,,
,1,2,
,1,2,
,k n i i S e e e a S n S S e i k i e i k i ∞⋅∞⋅∞⋅==、设多重集合=，表示集合满足下列条件的组合数，分别求数列的指数型生成函数:
(1)的每个元素出现奇数次；
(2)的每个元素至少出现4次；
（3）()至少出现次；（4）()至多出现次.
5、设有砝码重为1g 的3个，重为2g 的4个，重为4g 的2个，问能称出多少重量？各有几种方案？
6⨯、如果要把棋盘上偶数个方格涂成红色，试确定用红色、白色和蓝色对1n 棋盘的方格涂色的方法数。
组合数学(5)
当多重集S中的某个元素的个数为1时,则其 重复数可以省略。即
S= {2·a , 1·b , 3·c} = {2·a , b , 3·c} 如果S是一个多重集，那么S的一个r-排列 是S中r个元素的有序排放。如果S的元素总个 数是n(包括重复元素)，那么S的n-排列也称 为S的排列。对于 S= {2·a , 1·b , 3·c}
我们用“*”表示r-组合里的元素，用“”表示间
隔r-组合里各类型元素的分隔符，那么就可 以构造新的多重集T={r·* , (k-1)·}, 它只有两种 元素，它的排列数就等于S的r-组合数。由P41关 于两类型元素的排列数的求法得：
若进一步将8个车设为是互不相同的(比如,有8种 不同颜色的车),则在确定了8!种行方案后,对每 一种方案还要确定究竟是哪种颜色的车在哪方 格中。观察第一行到第八行,可以看到8种颜色 的一个排列。注意到对8!种位置排列的任一种 颜色的排列都有8!个,由乘法原理,在8×8棋盘上 具有8种不同颜色的8个非攻击型车的放置方案 数为: 8!·8!=(8!)2
把这些符号放到那些位置中的每一种方法决定 了一个选择方式。从“*”到第一个“”的数目
代表选择了m1个a1。在第一个“”和第二个“” 之间的数目m2代表了选择了m2个a2，等等。
因为选择“”的位置有:
种方法，因此有:
种选择方式，
它也等于选择“*”的位置的方法数:
因此，从S中允许重复的选择r元素无序 r-组合数为:
因此，可以把 C( n, n1) 看作是n个元素集合 的n1 - 组合数，也可以看成是具有两种类型 的元素并且它们的重复数分别是 n1和 n- n1的 多重集的排列的个数，此时排列与组合类似。
例求8×8棋盘上的8个非攻击型车的不同放法数 解：所谓两个车可以互相攻击，当且仅当二车 位于棋盘的同一行或同一列上。因此，非攻击 型车是指这些车占据着棋盘上的一些方格，
(完整word版)组合数学习题解答
第一章：1。
2. 求在1000和9999之间各位数字都不相同,而且由奇数构成的整数个数。
解:由奇数构成的4位数只能是由1,3，5，7，9这5个数字构成,又要求各位数字都不相同，因此这是一组从5个不同元素中选4个的排列，所以，所求个数为：P （5,4）=120。
1.4。
10个人坐在一排看戏有多少种就坐方式？如果其中有两人不愿坐在一起，问有多少种就坐方式？ 解:这显然是一组10个人的全排列问题，故共有10！种就坐方式。
如果两个人坐在一起，则可把这两个人捆绑在一起，如是问题就变成9个人的全排列，共有9!种就坐方式.而这两个人相捆绑的方式又有2种（甲在乙的左面或右面）。
故两人坐在一起的方式数共有2＊9！，于是两人不坐在一 起的方式共有 10！— 2＊9！.1.5. 10个人围圆桌而坐,其中两人不愿坐在一起，问有多少种就坐方式？ 解:这是一组圆排列问题，10个人围圆就坐共有10!10 种方式。
两人坐在一起的方式数为9!92⨯,故两人不坐在一起的方式数为：9！—2＊8！。
1。
14. 求1到10000中,有多少正数，它的数字之和等于5？又有多少数字之和小于5的整数？ 解：（1）在1到9999中考虑，不是4位数的整数前面补足0, 例如235写成0235,则问题就变为求: x 1+x 2+x 3+x 4=5 的非负整数解的个数，故有F （4,5）=⎪⎪⎭⎫⎝⎛-+=515456 （2）分为求：x 1+x 2+x 3+x 4=4 的非负整数解，其个数为F （4，4）=35 x 1+x 2+x 3+x 4=3 的非负整数解，其个数为F(4,3）=20 x 1+x 2+x 3+x 4=2 的非负整数解,其个数为F （4,2)=10 x 1+x 2+x 3+x 4=1 的非负整数解,其个数为F （4，1）=4 x 1+x 2+x 3+x 4=0 的非负整数解，其个数为F （4，0）=1将它们相加即得，F （4，4)+F(4，3）+F （4,2）+F （4，1）+F （4,0）=70。