Richard_A[1].Brualdi_组合数学习题解答
3.2组合数及其性质教案-2023-2024学年高二上学期数学北师大版(2019)选择性必修第一册
6. 教学指导书：为学生提供一本教学指导书，其中包括本节课的学习目标、教学内容、学习方法、练习题和答案等，以便学生能够更好地学习本节课的内容。
作用和目的：通过拓展练习和数学活动，帮助学生深入理解和应用组合数及其性质，提高学生的数学应用能力。
拓展与延伸
1. 提供与本节课内容相关的拓展阅读材料：
《组合数学导论》（作者：Richard A. Brualdi，ISBN：978-0-521-57498-1）
《概率论与数理统计》（作者：李尚志，ISBN：978-7-04-038045-0）
3.2组合数及其性质教案-2023-2024学年高二上学期数学北师大版（2019）选择性必修第一册
主备人
备课成员
教学内容分析
本节课的主要教学内容为组合数及其性质，属于北师大版（2019）选择性必修第一册，第3.2节。
组合数是组合数学中的基本概念，它是指从n个不同元素中，任取m（m≤n）个元素的所有不同组合的数目，记为C(n,m)。组合数具有以下性质：
板书设计
1. 组合数的定义与计算公式
- 定义：组合数是从n个不同元素中，任取m（m≤n）个元素的所有不同组合的数目，记为C(n,m)。
- 计算公式：C(n,m) = n! / [m!(n-m)!]。
2. 组合数的性质
- 非负性：C(n,m) ≥ 0，且当n=m时，C(n,m)=1。
- 对称性：C(n,m) = C(n,n-m)。
2. 组合数的性质：组合数的非负性、对称性和组合性。
3. C(5,2) = 5! / [2!(5-2)!] = 10。
应用组合数学(答案)
6. 对于图 A1.2a 的 Venn 图, 用每个区域所代表的集合的集合表达式来对该区域赋标签 (例如, 两个集 合的交的区域将会被标签为 A ∩ B).
7. 画出下列各集合的 Venn 图. (a) A − B. (b) A ∪ B. (c) (A ∪ B) ∩ (A ∩ B). (d) A − (B − A).
A.1 集 合 论 331
集中每一个都有两台机器. 参见图 A1.1. 那么选出 4 台机器的理想组合的策略之一是从图 A1.1 中的每一 个范畴中选出一台机器 (2×2×2×2=16 种选择). 从图 A1.1 中选出 4 台机器还有其他两种办法 (参见本节 末尾的练习 4).
涉及前述集合操作以及相关规则的集合表达式的研究称为布尔代数(Boolean algebra). 布尔代数的 3 个最重要的规则是
下面的集合表达式.
(a) (A ∪ B) ∩ (A ∩ B). (c) (A ∪ C) ∩ (A ∪ B) ∩ (B ∪ C).
(b) A − (B ∪ A).
(d) (A ∪ B) ∩ [(C ∪ B) − (A ∩ B)].
16. 假设投掷两枚骰子. 至少在一枚骰子上出现 1 或 2 的结果有多少种？
(b) {6, 10}. (d) {2, 7, 9}. (f ) {3, 5, 6, 7, 9, 10}.
3. 假设在例 1 中, 我们只知道这样的一条信息：两台计算器有内存但不可充电. 现在说明我们能够推出 图 A1.1 中其他 3 个盒子里的机器数量.
Eulid辗转相除法与二部图的一个对应
3. 正文
3.1. 一道组合最值问题
有若干正整数，它们和为 m ⋅ n ，且既可以分为和相等的 m 个组，又可以分为和相等的 n 个组，求这
些正整数个数的最小值 f (m, n) 。 经过对一些较小的 m 和 n 的试验我们猜测 f (m, n) 可能是 m + n − (m, n) 。我们的证明如下：
DOI: 10.12677/pm.2021.112030
221
理论ቤተ መጻሕፍቲ ባይዱ学
Eulid辗转相除法与二部图的一个对应
金长松，陈 贺 东北大学秦皇岛分校数学与统计学院，河北 秦皇岛
收稿日期：2021年1月2日；录用日期：2021年2月2日；发布日期：2021年2月9日
摘要
本文对一道组合最值问题的更一般情况进行了猜测和证明，利用了图论方法证明满足要求的值不小于目 标值，再利用归纳构造证明目标值是可行的。以此得到了Eulid辗转相除法与二部图的一个对应。
∑ ∑ x|(m,n)
m
+ x
n
−1
=
m + n − 1 ≥ m + n − (m, n)
x|(m,n) x
DOI: 10.12677/pm.2021.112030
220
理论数学
金长松，陈贺
由于图 G 的连线方法，如果两个点所代表的组都含有 f (m, n) ≥ m + n − (m, n) 中的某个数，则它们之
的点，其所在组里的正整数必定出现两次，否则一定可以再加进一个新的点与图 T 连通，与 G'的最大性
矛盾。对这些数求和，得到 ms = nt，于是其中点的个数为 s + t =m + n ，这里 x | (m, n) 因为图 G'是连通
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。
《组合数学》第五版 第6章答案.pdf
set size
justification
S
13 4
13 = 14 − 5 + 5 − 1
Ai
8 4
13 − 5 = 8
Ai ∩ Aj 0 13 − 5 − 5 = 3 < 4
By inclusion/exclusion
13
8
|A1 ∩ A2 ∩ A3 ∩ A4 ∩ A5| =
4
− 5 = 365. 4
set size
justification
S
17 3
17 = 14 + 4 − 1
Ai
8 3
17 − 9 = 8
Ai ∩ Aj 0 17 − 9 − 9 = −1 < 3
By inclusion/exclusion
17
8
|A1 ∩ A2 ∩ A3 ∩ A4| =
3
− 4 = 456. 3
8. Let S denote the set of positive integral solutions for x1 + x2 + x3 + x4 + x5 = 14. For 1 ≤ i ≤ 5 let Ai denote the set of elements in S with xi ≥ 6. We seek |A1 ∩A2 ∩A3 ∩A4 ∩A5|. We have
X∩Y ∩Z
0
15 − 5 − 4 − 5 = 1 < 3
X∩Y ∩W 0
15 − 5 − 4 − 6 = 0 < 3
X ∩Z ∩W
0 15 − 5 − 5 − 6 = −1 < 3
Y ∩Z∩W
0
抽屉原理在数学解题中的应用_吕松涛
［ 责任编辑 The Significance and Te aching Me thods to Start Mathe matics Mode lling Curriculum In Vocational Colle ge s
ZHU Fu，CHENG Yuan － an
（ Shangqiu Vocational and Technical College，Shangqiu 476000 ， China）
· 15·
2010 年
商丘职业技术学院学报
a2 ， a3 ， …， an ， a n +1 中必有相等的． 分析与证明： 设 G 为 n 阶有限群， 任取 a ∈ G ， 则由抽屉原理可知在 a， s t 1 t ＜ s n + 1 于是有 a s －t = e， 不妨设 a = a ， 从而 a 的阶有限．
2
2． 1
抽屉原理的应用
解决代数学问题 代数学中的一些问题非常的抽象或复杂 ， 解答起来比较困难， 但对于一些问题利用抽屉原理解决会起
到好的收效． 例 1 ． 证明： 有限群中每个元素的阶均有限．
收稿日期： 2010 － 12 － 18 基金项目： 河南省高校科技创新人才支持计划（ 2008HASTIT025 ） 作者简介： 吕松涛（ 1978 — ） ， 男， 河南杞县人， 商丘师范学院数学系助教， 硕士， 主要从事组合数学与图论研究。
孙胜利］
Abstract ： According to the current quality of Mathematics teaching in Vocational colleges ， we concluded that to start mathematics modelling curriculum is the inevitable trend of the reform of mathematics teaching finally discussed teaching method of mathematics modelling curriculum Ke y words ： mathematical modeling； mathematics in vocational colleges； teaching methods
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)
=
组合数学第四版（RichardA.Brualdi著）机械工业出版社课后答案1
第2章 鸽巢原理2.4 练习题1、关于本节中的应用4，证明对于每一个1，2，…，21存在连续若干天，在此期间国际象棋大师将恰好下完局棋（情形21是在应用4中处理的情况）。
能否判断：存在连续若干天，在此期间国际象棋大师将恰好下完22局棋？=k k =k 证明：设表示在前天下棋的总数i a i 若正好有=,则命题得证。
若不然,如下：i a k ∵共有11周，每天至少一盘棋，每周下棋不能超过12盘∴有 ，且771≤≤i 13217721≤<<<≤a a a{}21,,2,1 ∈∀k 有k k a k a k a k +≤+<<+<+≤+13217721观察以下154个整数：k a k a k a a a a +++77217721,,,,,,,每一个数是1到之间的整数，其中k +132153132≤+k由鸽巢原理，这154个数中至少存在两个相等的数∵都不相等，7721,,,a a a k a k a k a +++7721,,, 都不相等∴j i ,∃，使=i a k a j +即这位国际象棋大师在第，1+j 2+j ，…，天总共下了盘棋。
i k 综上所述，对于每一个1，2，…，21存在连续若干天，在此期间国际象棋大师将恰好下完局棋。
=k k □当=22时，132+=154，那么以下154个整数k k 22,,22,22,,,,77217721+++a a a a a a在1到154之间。
ⅰ）若这154个数都不相同则它们能取到1到154的所有整数，必然有一个数是22∵，2222>+i a 771≤≤i∴等于22的数必然是某个，i a 771≤≤i则在前天，这位国际象棋大师总共下了22盘棋。
i ⅱ）若这154个数中存在相同的两个数∵都不相等，7721,,,a a a k a k a k a +++7721,,, 都不相等∴j i ,∃，使=i a k a j +即这位国际象棋大师在第，1+j 2+j ，…，天总共下了盘棋。
组合数学(Richard A. Brualdi)重要知识考点
第一章 什么是组合§1 排列组合组合数学主要研究有限集合的计数，结构的存在性，以及性质。
几个计数原理设A 是有限多个元素的集合，用A 表示A 的元素个数a) 分类与加法原理设12=r A A A A …，i j A A =∅ ，i j ≠，则称{}1i A i r ≤≤为A 的一个分类，显然1=rii A A=∑（加法原理）b) 分步骤乘法原理设12r A A A 、、…是有限集令{}12123=(,,)|,1r r i i B A A A a a a a a A i r ⨯⨯⨯=∈≤≤…，…，（笛卡尔乘积），称B 为12r A A A ，，…，的积，显然B 的每个元素123(,,)r a a a a ，…，是有序数对，是按步骤确定的且1=rii B A=∏（乘法原理）【例1】{}{}121,2,3,4,A A ==则(){}121,3,(2,4),(1,4),(2,4)A A ⨯= 【例2】求120的因数的个数解：3120=235⨯⨯，312|120235aaan n ∴⇔=⋅⋅，其中12303,01,01a a a ≤≤≤≤≤≤ 令{}{}{}1230,1,2,3,0,1,0,1A A A ===，记120的因数的集合为B 故123||=||=422=16B A A A ⨯⨯⨯⨯.【例3】有限集{}123n a a a a ， ， ，…，的一个有序列123r i i i i a a a a ， ， ，…,称为123n a a a a ， ， ，…，的一个r 排列，其中i j a a ≠，i j ≠，所有r 排列的个数记为(,)(1)(1)P n r n n n r =--+……，令(,)!P n n n =，则!(,)()!n P n r n r =-，规定：0！=1.c) 双射原理设A 、B 是两个集合，对应:f A B →，对A 中的每个元素a ，有唯一的元素()b f a =，a D ∈，与之对应，则称f 为一个映射.映射：srr→→不是映射.1. 如果1212,(),a a A a a ∀∈≠有12()()f a f a ≠，则称f 是单射.显然，这时有A B ≤.2. 如果,b B ∀∈有,a A ∈使(),f a b =则称f 为满射.3. 如果f 既是单射又是满射，则称f 为双射.这时=A B .【例4】设{}123r A a a a a = ， ， ，…，，计算A 的所有子集的个数（组合证明） 解：设B 表示A 的所有子集所成的子集（或者用幂集2A表示） 设f ：{}{}{}{}0,10,10,10,1r B →⨯⨯⨯个……令x B ∈，123={}t i i i i x a a a a ， ， ，…,，1t r ≤≤，=0t 时表示∅，121t i i r ≤≤≤≤≤…i 12()(0000),t i i i f x =⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅111如果=0x ，则令()(000)f x =⋅⋅⋅⋅⋅⋅ 易知f 是双射.由双射原理和乘法原理得：=2r B 补充：例如{}1,2A =，{}{1}{2}{1,2}B =∅，，，在上述映射下，有={0,1}{0,1}B ⨯，()(0,0)f ∅=，({1})(1,0)f =，({2})(0,1)f = 【例5】{}123n a a a a ， ， ，…，的r 个元素所成的集合成为A 的一个r 组合所有这样的r 组合的个数记为(,)!n P n r r r ⎛⎫= ⎪⎝⎭，称之为二项式系数.一般来讲0r ≥，特别的当=0r 时，10n ⎛⎫= ⎪⎝⎭，今后，令(1)(1)!x x x x r r r ⎛⎫-⋅⋅⋅-+= ⎪⎝⎭，0r ≥且x 为任意实数. 注意12！是没有意义的.【例6】 1)!(0)!()!n n n n r r r n r ⎛⎫=≥≥ ⎪-⎝⎭为正整数， 2)()n n r n r ⎛⎫⎛⎫= ⎪ ⎪-⎝⎭⎝⎭对称性 3)111n n n r r r --⎛⎫⎛⎫⎛⎫=+ ⎪ ⎪ ⎪-⎝⎭⎝⎭⎝⎭ 4) ++201nn n n n ⎛⎫⎛⎫⎛⎫= ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭……循环排列（圆排列）：从{}123n a a a a ， ， ，…，中取出r 个元素作圆排列的个数为(,)!=()!P n r n r r n r -，当n r =时，!(1)!n n n =-如：1,2,3三个元素作圆排列，一共有3!23=种不同的排列方法.§3 重集设123k a a a a ， ， ，…，是个不同的元素，我们用{}1122k k n a n a n a ⋅⋅⋅ ， ，…，表示有1122k k n a n a n a 个 ，个 ，…，个的集合称为一个重集，这里0(1)i n i k ≤≤∞≤≤ 注：=0i n 表示i a 不出现，=i n ∞表示i a 取之不尽这个集合的r 元子集称为一个r 组合，r 元有序集合称为一个r 排列.【例1】{}123231a a a ⋅⋅⋅ ， ，的5组合有：{}1223a a ⋅⋅ ，、{}12321a a a ⋅⋅⋅ ，2 ，、{}123131a a a ⋅⋅⋅ ，，3个；6排列共有6!60231=⋅⋅！！！个.§4 重集的排列a){}12k a a a ∞⋅∞⋅∞⋅ ， ，…，的n 排列的个数等同于{}12k n a n a n a ⋅⋅⋅ ， ，…，的n 排列的个数=n kb){}1122k k n a n a n a ⋅⋅⋅ ， ，…，的全排列（这里1i n ≤≤∞）的个数为1212!k k n n n n n n n n ⎛⎫ ⎪⎝⎭…！！…！，其中1ki i n n ==∑ 注：12k n n n n ⎛⎫⎪⎝⎭…称为多项式系数，这里只有当12k n n n n =+++…时才有意义.证明：要得到{}1122k k n a n a n a ⋅⋅⋅ ， ，…，一个n 排列，我们只需从n 个有序位置中选取1n 个位置来放1a ，共有1n n ⎛⎫⎪⎝⎭种方法，再在余下的1n n -个位置中选2n 个位置来放2a ，共有12n n n -⎛⎫⎪⎝⎭种方法，继续下去，由乘法原理，n 排列的个数等于12121131212111211212312312112)!()!()!!!()!!()!!()!()!!k k k k k k k n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n ----------⎛⎫⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪⎪⎝⎭⎝⎭⎝⎭⎝⎭-------=⋅⋅-----------=………（…………！！…！c) 多重集合的组合。
抽屉原理的构造及其应用
抽屉原理的构造及其应用摘要：抽屉原理是组合数学中最基本的计数原理之一，是处理存在性问题的一个重要方法，本文主要介绍抽屉原理的几个构造方法以及一些应用。
关键词：抽屉原理、构造、应用。
抽屉原理将m个物品放入n个抽屉，则至少有一个抽屉中的物品个数不少于[（m-1/n]+1个，其中m-1/n向下取整由此可知，在利用抽屉原理解题时，首先要明确哪些是“物品”，哪些是“抽屉”，而这两者通常不会现存于题目中，尤其是抽屉，这个往往需要我们用一些巧妙的方法去构造，下面举例说明常见的抽屉构造方法。
1、利用分割图形的方法构造抽屉本方法主要用于解决点在几何图形中的位置分布和性质问题，通常我们把一个几何图形分割成几部分，然后把每一部分当做一个“抽屉”，每个抽屉里放入相应的元素．通常情况下，我们分割图形构造抽屉的最好方法是等分这个几何图形。
例1：从边长为2的正方形中任选5个点，则它们当中存在两个点，这两个点之间的距离至多为1.4. 证明：将此正方形分割成4个边长为1的小正方形。
当有两点位于其中一个小正方形时，这两个点之间的距离不会超过小正方形对角线的长度1.4.由抽屉原理知，5个点中必至少有两个点位于一个小正方形中。
2、利用划分数组的方法来构造抽屉利用此方法解题的关键是要明确分组的“对象”，然后将这些对象分成适当的数组。
再应用抽屉原理，问题便得以解决。
例2：由小于100的27个不同的奇数组成的集合中，必有两个数其和为102. 证明：将小于100的奇数分为26个组（抽屉）：{3,99}、{5,97}…{49,53}因为有27个奇数，把它看作物品，由抽屉原理可知，必有两个奇数落在同一抽屉中，这两个数之和恰好为102.例3：任意给定7个不同的整数，求证：其中必有两数之和或差是lO的倍数．证明：设这7个不同的整数分别为，a0…a，它们分别除以10后。
得到的余数是从0到9中的数．(1)若这7个余数中有两个数相同：设ai=10p+k，aj=10q+k(p、q为整数)，则ai-aj=10（p-q)，即ai-aj是lO的倍数，即存在两数之差是10的倍数． (2)若这7个余数中任何两个都不同．由抽屉原理知，至少有一数被lO除余数为6、7、8、9之一．将余数为6的数与余数为4的数划为一组余数为7的数与余数为3的数划为一组，余数为8的数与余数为2的数划为一组，余数为9的数与余数为1的数划为一组。
组合数学【第5版】(英文版)第3章答案
Math475Text:Brualdi,Introductory Combinatorics5th Ed. Prof:Paul TerwilligerSelected solutions for Chapter31.For1≤k≤22we show that there exists a succession of consecutive days during which the grandmaster plays exactly k games.For1≤i≤77let b i denote the number of gamesplayed on day i.Consider the numbers{b1+b2+···+b i+k}76i=0∪{b1+b2+···+b j}77j=1.There are154numbers in the list,all among1,2,...,153.Therefore the numbers{b1+b2+···+b i+k}76i=0∪{b1+b2+···+b j}77j=1.are not distinct.Therefore there exist integers i,j(0≤i<j≤77)such that b i+1+···+b j=k.During the days i+1,...,j the grandmaster plays exactly k games.2.Let S denote a set of100integers chosen from1,2,...,200such that i does not divide j for all distinct i,j∈S.We show that i∈S for1≤i≤15.Certainly1∈S since 1divides every integer.By construction the odd parts of the elements in S are mutually distinct and at most199.There are100numbers in the list1,3,5,...,199.Therefore each of 1,3,5,...,199is the odd part of an element of S.We have3×5×13=195∈S.Therefore none of3,5,13,15are in S.We have33×7=189∈S.Therefore neither of7,9is in S.We have11×17=187∈S.Therefore11∈S.We have shown that none of1,3,5,7,9,11,13,15 is in S.We show neither of6,14is in S.Recall33×7=189∈S.Therefore32×7=63∈S. Therefore2×32×7=126∈S.Therefore2×3=6∈S and2×7=14∈S.We show10∈S. Recall3×5×13=195∈S.Therefore5×13=65∈S.Therefore2×5×13=130∈S. Therefore2×5=10∈S.We now show that none of2,4,8,12are in S.Below we list the integers of the form2r3s that are at most200:1,2,4,8,16,32,64,128,3,6,12,24,48,96,192,9,18,36,72,144,27,54,108,81,162.In the above array each element divides everything that lies to the southeast.Also,each row contains exactly one element of S.For1≤i≤5let r i denote the element of row i that is contained in S,and let c i denote the number of the column that contains r i.We must have c i<c i−1for2≤i≤5.Therefore c i≥6−i for1≤i≤5.In particular c1≥5so r1≥16,and c2≥4so r2≥24.We have shown that none of2,4,8,12is in S.By the above comments i∈S for1≤i≤15.3.See the course notes.4,5,6.Given integers n≥1and k≥2suppose that n+1distinct elements are chosen from{1,2,...,kn}.We show that there exist two that differ by less than k.Partition{1,2,...,nk}=∪ni=1S i where S i={ki,ki−1,ki−2,...,ki−k+1}.Among our n+1chosen elements,there exist two in the same S i.These two differ by less than k.17.Partition the set{0,1,...,99}=∪50i=0S i where S0={0},S i={i,100−i}for1≤i≤49,S50={50}.For each of the given52integers,divide by100and consider the remainder. The remainder is contained in S i for a unique i.By the pigeonhole principle,there exist two of the52integers for which these remainders lie in the same S i.For these two integers the sum or difference is divisible by100.8.For positive integers m,n we consider the rational number m/n.For0≤i≤n divide the integer10i m by n,and call the remainder r i.By construction0≤r i≤n−1.By the pigeonhole principle there exist integers i,j(0≤i<j≤n)such that r i=r j.The integer n divides10j m−10i m.For notational convenience define =j−i.Then there exists a positive integer q such that nq=10i(10 −1)m.Divide q by10 −1and call the remainder r. So0≤r≤10 −2.By construction there exists an integer b≥0such that q=(10 −1)b+r. Writingθ=m/n we have10iθ=b+r 10 −1=b+r10+r102+r103+···Since the integer r is in the range0≤r≤10 −2this yields a repeating decimal expansion forθ.9.Consider the set of10people.The number of subsets is210=1024.For each subset consider the sum of the ages of its members.This sum is among0,1,...,600.By the pigeonhole principle the1024sums are not distinct.The result follows.Now suppose we consider at set of9people.Then the number of subsets is29=512<600.Therefore we cannot invoke the pigeonhole principle.10.For1≤i≤49let b i denote the number of hours the child watches TV on day i.Consider the numbers{b1+b2+···+b i+20}48i=0∪{b1+b2+···+b j}49j=1.There are98numbers in the list,all among1,2,...,96.By the pigeonhole principle the numbers{b1+b2+···+b i+20}48i=0∪{b1+b2+···+b j}49j=1.are not distinct.Therefore there existintegers i,j(0≤i<j≤49)such that b i+1+···+b j=20.During the days i+1,...,j the child watches TV for exactly20hours.11.For1≤i≤37let b i denote the number of hours the student studies on day i.Considerthe numbers{b1+b2+···+b i+13}36i=0∪{b1+b2+···+b j}37j=1.There are74numbers in the list,all among1,2,...,72.By the pigeonhole principle the numbers{b1+b2+···+b i+13}36i=0∪{b1+b2+···+b j}37j=1are not distinct.Therefore there exist integers i,j(0≤i<j≤37)such that b i+1+···+b j=13.During the days i+1,...,j the student will have studied exactly13hours.12.Take m=4and n=6.Pick a among0,1,2,3and b among0,1,2,3,4,5such that a+b is odd.Suppose that there exists a positive integer x that yields a remainder of a(resp.b) when divided by4(resp.by6).Then there exist integers r,s such that x=4r+a and x=6s+bining these equations we obtain2x−4r−6s=a+b.In this equation the2left-hand side is even and the right-hand side is odd,for a contradiction.Therefore x does not exist.13.Since r (3,3)=6there exists a K 3subgraph of K 6that is red or blue.We assume that this K 3subgraph is unique,and get a contradiction.Without loss we may assume that the above K 3subgraph is red.Let x denote one of the vertices of this K 3subgraph,and let {x i }5i =1denote the remaining five vertices of K 6.Consider the K 5subgraph with vertices{x i }5i =1.By assumption this subgraph has no K 3subgraph that is red or blue.The only edge coloring of K 5with this feature is shown in figure 3.2of the text.Therefore we may assume that the vertices {x i }5i =1are labelled such that for distinct i,j (1≤i,j ≤5)the edge connecting x i ,x j is red (resp.blue)if i −j =±1modulo 5(resp.i −j =±2modulo5).By construction and without loss of generality,we may assume that each of x 1,x 2is connected to x by a red edge.Thus the vertices x,x 1,x 2give a red K 3subgraph.Now the edge connecting x and x 3is blue;otherwise the vertices x,x 2,x 3give a second red K 3subgraph.Similarly the edge connecting x and x 5is blue;otherwise the vertices x,x 1,x 5give a second red K 3subgraph.Now the vertices x,x 3,x 5give a blue K 3subgraph.14.After n minutes we have removed n pieces of fruit from the bag.Suppose that among the removed fruit there are at most 11pieces for each of the four kinds.Then our total n must be at most 4×11=44.After n =45minutes we will have picked at least a dozen pieces of fruit of the same kind.15.For 1≤i ≤n +1divide a i by n and call the reminder r i .By construction 0≤r i ≤n −1.By the pigeonhole principle there exist distinct integers i,j among 1,2,...,n +1such that r i =r j .Now n divides a i −a j .bel the people 1,2,...,n .For 1≤i ≤n let a i denote the number of people aquainted with person i .By construction 0≤a i ≤n −1.Suppose the numbers {a i }n i =1are mutually distinct.Then for 0≤j ≤n −1there exists a unique integer i (1≤i ≤n )such that a i =j .Taking j =0and j =n −1,we see that there exists a person aquainted with nobody else,and a person aquainted with everybody else.These people are distinct since n ≥2.These two people know each other and do not know each other,for a contradiction.Therefore the numbers {a i }n i =1are not mutually distinct.17.We assume that the conclusion is false and get a bel the people 1,2,...,100.For 1≤i ≤100let a i denote the number of people aquainted with person i .By construction 0≤a i ≤99.By assumption a i is even.Therefore a i is among 0,2,4,...,98.In this list there are 50numbers.Now by our initial assumption,for each even integer j (0≤j ≤98)there exists a unique pair of integers (r,s )(1≤r <s ≤100)such that a r =j and a s =j .Taking j =0and j =98,we see that there exist two people who know nobody else,and two people who know everybody else except one.This is a contradiction.18.Divide the 2×2square into four 1×1squares.By the pigeonhole principle there exists a 1×1square that contains at least two of the five points.For these two points the distance apart is at most √2.319.Divide the equilateral triangle into a grid,with each piece an equilateral triangle of side length1/n.In this grid there are1+3+5+···+2n−1=n2pieces.Suppose we place m n=n2+1points within the equilateral triangle.Then by the pigeonhole principle there exists a piece that contains two or more points.For these two points the distance apart is at most1/n.20.Color the edges of K17red or blue or green.We show that there exists a K3subgraph of K17that is red or blue or green.Pick a vertex x of K17.In K17there are16edges that contain x.By the pigeonhole principle,at least6of these are the same color(let us say red).Pick distinct vertices{x i}6i=1of K17that are connected to x via a red edge.Consider theK6subgraph with vertices{x i}6i=1.If this K6subgraph contains a red edge,then the twovertices involved together with x form the vertex set of a red K3subgraph.On the other hand,if the K6subgraph does not contain a red edge,then since r(3,3)=6,it contains a K3subgraph that is blue or green.We have shown that K17has a K3subgraph that is red or blue or green.21.Let X denote the set of sequences(a1,a2,a3,a4,a5)such that a i∈{1,−1}for1≤i≤5 and a1a2a3a4a5=1.Note that|X|=16.Consider the complete graph K16with vertex set X.We display an edge coloring of K16with colors red,blue,green such that no K3 subgraph is red or blue or green.For distinct x,y in X consider the edge connecting x and y.Color this edge red(resp.blue)(resp.green)whenever the sequences x,y differ in exactly 4coordinates(resp.differ in exactly2coordinates i,j with i−j=±1modulo5)(resp. differ in exactly2coordinates i,j with i−j=±2modulo5).Each edge of K16is now colored red or blue or green.For this edge coloring of K16there is no K3subgraph that is red or blue or green.22.For an integer k≥2abbreviate r k=r(3,3,...,3)(k3’s).We show that r k+1≤(k+1)(r k−1)+2.Define n=r k and m=(k+1)(n−1)+2.Color the edges of K m with k+1colors C1,C2,...,C k+1.We show that there exists a K3subgraph with all edges the same color.Pick a vertex x of K m.In K m there are m−1edges that contain x.By the pigeonhole principle,at least n of these are the same color(which we may assume is C1).Pick distinct vertices{x i}ni=1of K m that are connected to x by an edge colored C1.Considerthe K n subgraph with vertices{x i}ni=1.If this K n subgraph contains an edge colored C1,then the two vertices involved together with x give a K3subgraph that is colored C1.On the other hand,if the K n subgraph does not contain an edge colored C1,then since r k=n, it contains a K3subgraph that is colored C i for some i(2≤i≤k+1).In all cases K m hasa K3subgraph that is colored C i for some i(1≤i≤k+1).Therefore r k≤m.23.We proved earlier thatr(m,n)≤m+n−2n−1.Applying this result with m=3and n=4we obtain r(3,4)≤10.24.We show that r t(t,t,q3)=q3.By construction r t(t,t,q3)≥q3.To show the reverse inequality,consider the complete graph with q3vertices.Let X denote the vertex set of this4graph.Color the t -element subsets of X red or blue or green.Then either (i)there exists a t -element subset of X that is red,or (ii)there exists a t -element subset of X that is blue,or (iii)every t -element subset of X is green.Therefore r t (t,t,q 3)≤q 3so r t (t,t,q 3)=q 3.25.Abbreviate N =r t (m,m,...,m )(k m ’s).We show r t (q 1,q 2,...,q k )≤N .Consider the complete graph K N with vertex set X .Color each t -element subset of X with k colors C 1,C 2,...,C k .By definition there exists a K m subgraph all of whose t -element subsets are colored C i for some i (1≤i ≤k ).Since q i ≤m there exists a subgraph of that K m with q i vertices.For this subgraph every t -element subset is colored C i .26.In the m ×n array assume the rows (resp.columns)are indexed in increasing order from front to back (resp.left to right).Consider two adjacent columns j −1and j .A person in column j −1and a person in column j are called matched if they occupy the same row of the original formation.Thus a person in column j is taller than their match in column j −1.Now consider the adjusted formation.Let L and R denote adjacent people in some row i ,with L in column j −1and R in column j .We show that R is taller than L.We assume that L is at least as tall as R,and get a contradiction.In column j −1,the people in rows i,i +1,...,m are at least as tall as L.In column j ,the people in rows 1,2,...,i are at most as tall as R.Therefore everyone in rows i,i +1,...,m of column j −1is at least as tall as anyone in rows 1,2,...,i of column j .Now for the people in rows 1,2,...,i of column j their match stands among rows 1,2,...,i −1of column j −1.This contradicts the pigeonhole principle,so L is shorter than R.27.Let s 1,s 2,...,s k denote the subsets in the collection.By assumption these subsets are mutually distinct.Consider their complements s 1,s 2,...,s k .These complements are mutu-ally distinct.Also,none of these complements are in the collection.Therefore s 1,s 2,...,s k ,s 1,s 2,...,s k are mutually distinct.Therefore 2k ≤2n so k ≤2n −1.There are at most 2n −1subsets in the collection.28.The answer is 1620.Note that 1620=81×20.First assume that 100i =1a i <1620.We show that no matter how the dance lists are selected,there exists a group of 20men that cannot be paired with the 20women.Let the dance lists be bel the women 1,2,...,20.For 1≤j ≤20let b j denote the number of men among the 100that listed woman j .Note that 20j =1b j = 100i =1a i so ( 20j =1b j )/20<81.By the pigeonhole principle there exists an integer j (1≤j ≤20)such that b j ≤80.We have 100−b j ≥20.Therefore there exist at least 20men that did not list woman j .This group of 20men cannot be paired with the 20women.Consider the following selection of dance lists.For 1≤i ≤20man i lists woman i and no one else.For 21≤i ≤100man i lists all 20women.Thus a i =1for 1≤i ≤20and a i =20for 21≤i ≤100.Note that 100i =1a i =20+80×20=1620.Note also that every group of 20men can be paired with the 20women.29.Without loss we may assume |B 1|≤|B 2|≤···≤|B n |and |B ∗1|≤|B ∗2|≤···≤|B ∗n +1|.By assumption |B ∗1|is positive.Let N denote the total number of objects.Thus N = n i =1|B i |and N = n +1i =1|B ∗i |.For 0≤i ≤n define∆i =|B ∗1|+|B ∗2|+···+|B ∗i +1|−|B 1|−|B 2|−···−|B i |.5We have∆0=|B∗1|>0and∆n=N−N=0.Therefore there exists an integer r(1≤r≤n)such that∆r−1>0and∆r≤0.Now0<∆r−1−∆r=|B r|−|B∗r+1|so|B∗r+1|<|B r|.So far we have|B∗1|≤|B∗2|≤···≤|B∗r+1|<|B r|≤|B r+1|≤···≤|B n|.Thus|B∗i|<|B j|for1≤i≤r+1and r≤j≤n.Defineθ=|(B∗1∪B∗2∪···∪B∗r+1)∩(B r∪B r+1∪···∪B n)|.We showθ≥ing∆r−1>0we have|B∗1|+|B∗2|+···+|B∗r|>|B1|+|B2|+···+|B r−1|=|B1∪B2∪···∪B r−1|≥|(B1∪B2∪···∪B r−1)∩(B∗1∪B∗2∪···∪B∗r+1)|=|B∗1∪B∗2∪···∪B∗r+1|−θ=|B∗1|+|B∗2|+···+|B∗r+1|−θ≥|B∗1|+|B∗2|+···+|B∗r|+1−θ.Thereforeθ>1soθ≥2.6。
一笔画问题(欧拉图)
2010-10-18 17:32 by EricZhang(T2噬菌体), 3556 visits, 网摘, 收藏, 编辑关于一笔画问题的数学分析（对一道面试题的总结与扩展思考）摘要前几天参加了一个公司的面试，其中被问到了一个题。
面试官在纸上画了一个图形（具体图形见下文），问我能不能一笔画出这个图形，要求每条边必须只走一次，并且画的过程中笔不能离开纸。
当时我没有试着去画，而是凭着自己图论方面的知识在几秒钟之内告诉面试官不可能做到，然后简单说了一下理由。
面试结束后我翻阅了图论相关的资料，发现当时自己虽然给出了正确答案，但理由并不完全正确。
昨天我花了几个小时仔细研究了一下相关的理论，总结了一下这类问题的类型和解法，写成此文，分享给大家。
问题的提出当时面试官给我出的问题是这样的：对于下面这个图形，让我一笔画出，要求每条边必须只走一次，并且画的过程中笔不能离开纸。
面试时我给出的回答是不可能做到，面试结束后我也从数学上证明了这个这个回答。
当然有兴趣的朋友可以试着画画看。
这个问题其实就是我们小时候会玩到的一笔画游戏。
这类问题看似简单直观，但是仔细研究下来却蕴含了很多东西，而且涉及了图论中一个非常重要的研究课题——欧拉迹。
而且这类问题可以扩展出很多东西，例如任意给一个图可不可以完成一笔画且最后回到起始点？再如到底什么样的图可以一笔画出来？什么样的图一笔画不出来？如果一个图可以一笔画出来，那么应该如何画？有没有对一切可一笔画图形的通用解法？下面我们将这个问题抽象成一般问题，然后从图论角度寻找上述疑问的答案。
图论中的一些概念因为在下文论述过程中需要用到一些图论的基本概念，为了照顾在这方面不熟悉的朋友，我先将要用到的定义和概念列出来，如果您对图论的基本内容已经了然于胸，可以跳过这一节。
另外如不做特殊说明，下文所有的“图”都默认指“无向图”，本文的讨论不涉及“有向图”。
简单图——一个简单图可表示为G=(V, E)，其中V是顶点集合，其中每个元素是图的一个顶点；E是边集合，其中每一个的元素是一个顶点对(a, b)，其中a和b均属于V，这个顶点对表示顶点a和b 间有一条边相连。
组合数学 第四版 (Richard A[1].Brualdi 著) 机械工业出版社作业答案
![组合数学 第四版 (Richard A[1].Brualdi 著) 机械工业出版社作业答案](https://img.taocdn.com/s3/m/970ef108763231126edb11c4.png)
解法 1 因此，
对任意非负整数 n 和 k， (k 1) 即 ， ， (n 1) k k k 1 k 1 k 1 n 1 1 n 1 n 1 n 1 n 1 (1) n 2 1 3 2 4 3 n 1 n
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a1 , a 2 , , a 37 和 a1 13, a 2 13,, a 37 13 都是严格单调递增序列。因为总的学习时间
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③ 考虑 6 位整数。最高位不能为 0，因此 8 位整数有 7 P(7, 5) 个。 ④ 考虑 5 位整数。最高位不能为 0，因此 8 位整数有 7 P (7, 4) 个。 ⑤ 考虑 4 位整数。若千位数字大于 5，有 3 P(7, 3) 个。若千位数字等于 5，则百位数字 必须大于等于 4，有 4 P(6, 2) 个。
至少已拿出 12 个相同种类的水果。因此，需要 45 分钟。 17. 证明：在一群 n 1 个人中，存在两个人，他们在这群人中有相同数目的熟人（假设没 有人与他/她自己是熟人） 。
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证明 因为每个人都不是自己的熟人，所以每个人的熟人的数目是从 0 到 n 1 的整数。若 有两个人的熟人的数目分别是 0 和 n 1 ，则有人谁都不认识，有人认识所有的人，这是不 可能的。因此，这 n 个人的熟人的数目是 n 1 个整数之一，必有两个人有相同数目的熟人。
组合数学第五版第七章答案
1
0
1
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1
2
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The sequence repeats with period 3 (resp. 8) (resp. 6) if m = 2 (resp. m = 3) (resp. m = 4). The result follows.
4. For an integer n ≥ 5 we have
8. By construction h0 = 1 and h1 = 2. We now find hn for n ≥ 2. Consider a coloring of the 1 × n chessboard. The first square is colored red or blue. If it is blue, then there are hn−1 ways to color the remaining n − 1 squares. If it is red, then the second square is blue, and there are hn−2 ways to color the remaining n − 2 squares. Therefore hn = hn−1 + hn−2. Comparing the above data with the Fibonacci sequence we find hn = fn+2.
m = qn + r
1 ≤ r ≤ n − 1.
Observe that
GCD(n, r) = GCD(m, n) = d.
By induction and since r ≤ n − 1,
Richard组合数学第5版-第4章课后习题答案(英文版)
7365 412
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8. (a) For a permutation of {1, 2, 3, 4, 5, 6} the corresponding inversion sequence
(b1, b2, b3, b4, b5, b6) satisfies 0 ≤ bi ≤ 6 − i for 1 ≤ i ≤ 6. The total number of inversions is
←− ←− ←− ←− ←− 12534 ←− ←− →− ←− ←− 12543 ←− ←− ←− ←− ←− 14523 ←−4 ←−1 →−5 ←−2 ←−3 →−4 ←−1 ←−5 ←−3 ←−2 ←−1 →−4 →−5 ←−3 ←−2 ←− ←− ←− →− ←− 13542 ←− ←− →− ←− →− 13524 ←− ←− ←− ←− ←− 31524 ←−3 ←−1 →−5 ←−4 ←−2
(c) The set {x6} corresponds to coordinate 7 6 5 4 3 2 1 0 entry 0 1 0 0 0 0 0 0
《组合数学》测试题含答案
《组合数学》测试题含答案测试题——组合数学⼀、选择题1. 把101本书分给10名学⽣，则下列说法正确的是（）A.有⼀名学⽣分得11本书B.⾄少有⼀名学⽣分得11本书C.⾄多有⼀名学⽣分得11本书D.有⼀名学⽣分得⾄少11本书2. 8⼈排队上车，其中A ，B 两⼈之间恰好有4⼈，则不同的排列⽅法是（）A.!63?B.!64?C. !66?D. !68?3. 10名嘉宾和4名领导站成⼀排参加剪彩，其中领导不能相邻，则站位⽅法总数为（）A.()4,11!10P ? B. ()4,9!10P ? C. ()4,10!10P ? D. !3!14-4. 把10个⼈分成两组，每组5⼈，共有多少种⽅法（）A.???? ??510B.510510 C.49 D. 4949 5. 设x,y 均为正整数且20≤+y x ，则这样的有序数对()y x ,共有（）个A.190B.200C.210D.2206. 仅由数字1，2，3组成的七位数中，相邻数字均不相同的七位数的个数是（）A.128B.252C.343D.1927. 百位数字不是1且各位数字互异的三位数的个数为（）A.576B.504C.720D.3368. 设n 为正整数，则∑=???? ??nk k n 02等于（）A.n 2B. 12-n C. n n 2? D. 12-?n n 9. 设n 为正整数，则()k kn k k n 310???? ??-∑=的值是（）A.n 2B. n 2-C. ()n2- D.0 10. 设n 为正整数，则当2≥n 时，∑=???? ??-nk k k 22=()A.3n B. ?+21n C. +31n D. 22+ n 11. ()632132x x x +-中23231x x x 的系数是（）A.1440B.-1440C.0D.112. 在1和610之间只由数字1，2或3构成的整数个数为（） A.2136- B. 2336- C. 2137- D. 2337 - 13. 在1和300之间的整数中能被3或5整除的整数共有（）个A.100B.120C.140D.16014. 已知(){}o n n f ≥是Fibonacci 数列且()()348,217==f f ，则()=10f （） A.89 B.110 C.144 D.28815. 递推关系3143---=n n n a a a 的特征⽅程是（）A.0432=+-x xB. 0432=-+x xC. 04323=+-x xD. 04323=-+x x16. 已知()??=?+=,2,1,0232n a n n ，则当2≥n 时，=n a （）A.2123--+n n a aB. 2123---n n a aC.2123--+-n n a aD. 2123----n n a a17. 递推关系()=≥+=-312201a n a a n n n 的解为（） A.32+?=n n n a B. ()221+?+=n n n aC. ()122+?+=n n n aD. ()n n n a 23?+=18. 设()??=?=,2,1,025n a n n ，则数列{}0≥n n a 的常⽣成函数是（）A.x 215-B. ()2215x - C.()x 215- D. ()2215x -19. 把15个相同的⾜球分给4个⼈，使得每⼈⾄少分得3个⾜球，不同的分法共有（）种A.45B.36C.28D.2020. 多重集{}b a S ??=4,2的5-排列数为（）A.5B.10C.15D.2021. 部分数为3且没有等于1的部分的15-分拆的个数为（）A.10B.11C.12D.1322. 设n,k 都是正整数，以()n P k 表⽰部分数为k 的n-分拆的个数，则()116P 的值是（）A.6B.7C.8D.923. 设A ，B ，C 是实数且对任意正整数n 都有+ + =1233n C n B n A n ，则B 的值是（）A.9B.8C.7D.624. 不定⽅程1722321=++x x x 的正整数解的个数是（）A.26B.28C.30D.3225. 已知数列{}0≥n n a 的指数⽣成函数是()()t t e e t E 521?-=，则该数列的通项公式是（）A.n n n n a 567++=B. n n n n a 567+-=C. n n n n a 5627+?+=D. n n n n a 5627+?-=⼆、填空题1. 在1和2000之间能被6整除但不能被15整除的正整数共有_________个2. ⽤红、黄、蓝、⿊4种颜⾊去图n ?1棋盘，每个⽅格涂⼀种颜⾊，则使得被涂成红⾊的⽅格数是奇数的涂⾊⽅法共有_______种3. 已知递归推关系()31243321≥-+=---n a a a a n n n n 的⼀个特征根为2，则其通解为___________4. 把()3≥n n 个⼈分到3个不同的房间，每个房间⾄少1⼈的分法数为__________5. 棋盘??的车多项式为___________6. 由5个字母a,b,c,d,e 作成的6次齐次式最多可以有_________个不同类的项。
转论文男女配对婚姻模型
转论文男女配对婚姻模型摘要：本文针对现在社会中存在的诸多婚恋问题，建立关于配对和婚姻的数学模型，并且通过对模型的分析得出了惊人的结论！本文在针对婚恋问题进行了机理分析后，得出人们配对的方法和选择配偶的方法。
本文将选择配偶的生活问题数学化，并且建立了关于配对和稳定婚姻的数学模型，通过图论中二分图的相关知识，分析并解决了如何建立配对并且选择配偶保证稳定完备婚姻，并通过延迟认可算法得到了选择配偶的方式。
最终，通过对模型的分析得出惊人结论。
关键词：配对稳定完备婚姻二分图延迟认可算法一、问题重述：男女的婚嫁问题是关系到人类繁衍的重大问题。
每个人都希望找到自己心仪的人作为自己的配偶，而对于普通人来说，究竟是怎么可以进行配对并且结婚呢?同时，离婚、婚外情的不断发生也提醒人们：什么样的婚姻是稳定的呢?究竟怎么样才能得到一个稳定的婚姻呢?本文将通过建立数学模型讨论并解决这些问题。
二、模型的建立和求解：1.关于配对问题：2.1.1：问题描述有n位男士和m位女士，所有的男士都急于想结婚。
如果对娶谁没有限制，那么只要求女士人数m至少与男士人数n一样多就可以使所有男士结上婚。
但是，我们一般预料每一位男士会坚持与配偶的某种和谐从而会排除某些女士作为可能的配偶。
这样，每一位男士就会从这些女士中得到可接受的配偶的人选(包括女士也可以接受这名男士)，如果男士如果娶了自己可接受的配偶，就称为完备婚姻对应。
2.1.2：模型建立令Y为一有限集，令A=(A1,A2,.,An)为Y的n个子集的族(族和序列实际上是一样的。
这里的项是集合的序列。
与数列一样，不同的项可以相等；就是说，这些集不必是不同的。
)Y的元素的族(e1,e2,.,ei)叫做A的代表系统，如果e1在A1中，e2在A2中，.，en在An中。
在代表系统中，元素ei属于Ai从而"代表"Ai。
如果在代表系统中元素e1,e2,.,en是互异的，则(e1,e2,.,en)称为互异代表系统(system ofdistinct representatives)，简记为SDR。