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。