组合数学第三章习题解答

组合数学卢开澄课后习题答案组合数学是一门研究离散结构和组合对象的数学学科，它广泛应用于计算机科学、统计学、密码学等领域。

卢开澄是中国著名的组合数学家，他的教材《组合数学》是该领域的经典之作。

在学习组合数学的过程中，课后习题是巩固知识、提高能力的重要途径。

下面我将为大家提供一些卢开澄课后习题的答案。

第一章：集合与命题逻辑1.1 集合及其运算习题1：设集合A={1,2,3}，B={2,3,4}，求A∪B和A∩B的结果。

答案：A∪B={1,2,3,4}，A∩B={2,3}。

习题2：证明若A∩B=A∩C，且A∪B=A∪C，则B=C。

答案：首先，由A∩B=A∩C可得B⊆C，同理可得C⊆B，因此B=C。

然后，由A∪B=A∪C可得B⊆C，同理可得C⊆B，因此B=C。

综上所述，B=C。

1.2 命题逻辑习题1：将下列命题用命题变元表示：（1）如果今天下雨，那么我就带伞。

（2）要么他很聪明，要么他很勤奋。

答案：（1）命题变元P表示今天下雨，命题变元Q表示我带伞，命题可表示为P→Q。

（2）命题变元P表示他很聪明，命题变元Q表示他很勤奋，命题可表示为P∨Q。

习题2：判断下列命题是否为永真式、矛盾式或可满足式：（1）(P∨Q)→(P∧Q)（2）(P→Q)∧(Q→P)答案：（1）该命题为可满足式，因为当P为真，Q为假时，命题为真。

（2）该命题为永真式，因为无论P和Q取何值，命题都为真。

第二章：排列与组合2.1 排列习题1：从10个人中选取3个人，按照顺序排成一队，有多少种不同的结果？答案：根据排列的计算公式，共有10×9×8=720种不同的结果。

习题2：从10个人中选取3个人，不考虑顺序，有多少种不同的结果？答案：根据组合的计算公式，共有C(10,3)=120种不同的结果。

2.2 组合习题1：证明组合恒等式C(n,k)=C(n,n-k)。

答案：根据组合的计算公式可得C(n,k)=C(n,n-k)，因此组合恒等式成立。

课时作业(六) 组合数的应用一、选择题1．圆上有10个点，过每三个点画一个圆内接三角形，如此一共可以画的三角形个数为( )A．720 B．360C．240 D．1202．某电视台连续播放5个广告，其中有3个不同的商业广告和2个不同的奥运广告．要求最后必须播放奥运广告，且2个奥运广告不能连续播放，如此不同的播放方式有( )A．120种B．48种C．36种D．18种3．假如从1,2,3，…，9这9个整数中同时取4个不同的数，其和为偶数，如此不同的取法共有( )A．60种B．63种C．65种D．66种4．将标号为1,2，…，10的10个球放入标号为1,2，…，10的10个盒子里，每个盒内放一个球，恰好3个球的标号与其在盒子的标号不一致的放入方法种数为( )A．120 B．240C．360 D．720二、填空题5．某单位有15名成员，其中男性10人，女性5人，现需要从中选出6名成员组成考察团外出参观学习，如果按性别分层，并在各层按比例随机抽样，如此此考察团的组成方法种数是________．6．某球队有2名队长和10名队员，现选派6人上场参加比赛，如果场上最少有1名队长，那么共有________种不同的选法．7．现有6X风景区门票分配给6位游客，假如其中A，B风景区门票各2X，C，D风景区门票各1X，如此不同的分配方案共有________种．三、解答题8．某医院有内科医生12名，外科医生8名，现选派5名参加赈灾医疗队．(1)假如内科医生甲与外科医生乙必须参加，共有多少种不同选法？(2)假如甲、乙均不能参加，有多少种选法？(3)假如甲、乙2人至少有1人参加，有多少种选法？(4)假如医疗队中至少有1名内科医生和1名外科医生，有多少种选法？9．10件不同产品中有4件是次品，现对它们进展一一测试，直至找出所有4件次品为止．(1)假如恰在第5次测试，才测试到第一件次品，第10次才找到最后一件次品，如此这样的不同测试方法数是多少？(2)假如恰在第5次测试后，就找出了所有4件次品，如此这样的不同测试方法数是多少？[尖子生题库]10．按照如下要求，分别求有多少种不同的方法？(1)6个不同的小球放入4个不同的盒子；(2)6个不同的小球放入4个不同的盒子，每个盒子至少一个小球；(3)6个一样的小球放入4个不同的盒子，每个盒子至少一个小球．课时作业(六) 组合数的应用1．解析：确定三角形的个数为C310＝120.答案：D2．解析：最后必须播放奥运广告有C12种，2个奥运广告不能连续播放，倒数第2个广告有C13种，故共有C12C13A33＝36种不同的播放方式．答案：C3．解析：均为奇数时，有C45＝5种；均为偶数时，有C44＝1种；两奇两偶时，有C24·C25＝60种，共有66种．答案：D4．解析：先选出3个球有C310＝120种方法，不妨设为1,2,3号球，如此1,2,3号盒中能放的球为2,3,1或3,1,2两种．这3个放入标号不一致的盒子中有2种不同的方法，故共有120×2＝240种方法．答案：B5．解析：按性别分层，并在各层按比例随机抽样，如此需从10名男性中抽取4人，5名女性中抽取2人，共有C410C25＝2 100种抽法．答案：2 1006．解析：假如只有1名队长入选，如此选法种数为C12·C510；假如两名队长均入选，如此选法种数为C410，故不同选法有C12·C510＋C410＝714(种)．答案：7147．解析：6位游客选2人去A风景区，有C26种，余下4位游客选2人去B风景区，有C24种，余下2人去C，D风景区，有A22种，所以分配方案共有C26C24A22＝180(种)．答案：1808．解析：(1)只需从其他18人中选3人即可，共有C318＝816(种)选法．(2)只需从其他18人中选5人即可，共有C518＝8 568(种)选法．(3)分两类：甲、乙中有1人参加；甲、乙都参加．如此共有C12C418＋C318＝6 936(种)选法．(4)方法一(直接法)：至少有1名内科医生和1名外科医生的选法可分4类：1内4外；2内3外；3内2外；4内1外．所以共有C112C48＋C212C38＋C312C28＋C412C18＝14 656(种)选法．方法二：从无限制条件的选法总数中减去5名都是内科医生和5名都是外科医生的选法种数所得的结果即为所求，即共有C520－(C512＋C58)＝14 656(种)选法．9．解析：(1)先排前4次测试，只能取正品，有A46种不同测试方法，再从4件次品中选2件排在第5和第10的位置上测试，有C24A22＝A24种测法，再排余下4件的测试位置，有A44种测法．所以共有不同测试方法A46·A24·A44＝103 680种．(2)第5次测试恰为最后一件次品，另3件在前4次中出现，从而前4次有一件正品出现，所以共有不同测试方法C16·C34·A44＝576种．10．解析：(1)每个小球都有4种方法，根据分步乘法计数原理，共有46＝4 096种不同放法．(2)分两类：第1类，6个小球分3,1,1,1放入盒中；第2类，6个小球分2,2,1,1放入盒中，共有C36·C14·A33＋C26·C24·A24＝1 560(种)不同放法．(3)方法一：按3,1,1,1放入有C14种方法，按2,2,1,1，放入有C24种方法，共有C14＋C24＝10(种)不同放法．方法二：(挡板法)在6个球之间的5个空中插入三个挡板，将6个球分成四组，共有C35＝10(种)不同放法．。

★★★第一章：★★★1、用六种方法求839647521之后的第999个排列。

提示：先把999换算成递增或递减进位制数，加到中介数上，就不用计算序号了。

解：字典序法递增进位制法递减进位制法邻位对换法839647521的中介数72642321↑67342221↑12224376↓10121372↓999的中介数121211↑121211↑1670↓1670↓839647521后999的中介数73104210↑67504110↑12230366↓10123362↓839647521后999个的排列842196537 859713426 389547216 →3←8→4→5→7→6←9←21★★★第二章★★★例5：10个数字（0到9）和4个四则运算符(+,-,×,÷) 组成的14个元素。

求由其中的n个元素的排列构成一算术表达式的个数。

因所求的n个元素的排列是算术表达式，故从左向右的最后一个符号必然是数字。

而第n-1位有两种可能，一是数字，一是运算符。

如若第n-1位是十个数字之一，则前n-1位必然构成一算术表达式。

10a n-1如若不然，即第n-1位是4个运算符之一，则前n-2位必然是算术表达式。

40a n-2,根据以上分析，令a n表示n个元素排列成算术表达式的个数。

则a2=120指的是从0到99的100个数，以及±0,±1,...,±9,利用递推关系(2-8-1)，得a0=1/2特征多项式x2-10x-40 。

它的根是解方程得例7：平面上有一点P，它是n个域D1,D2,...,D n的共同交界点，见图2-8-4现取k种颜色对这n个域进行着色，要求相邻两个域着的颜色不同。

试求着色的方案数。

令a n表示这n个域的着色方案数。

无非有两种情况（1）D1和D n-1有相同的颜色；（2）D1和D n-1所着颜色不同。

第一种情形，域有k-1种颜色可用，即D1D n-1域所用颜色除外；而且从D1到D n-2的着色方案，和n-2个域的着色方案一一对应。

习题一（排列与组合）1．在1到9999之间，有多少个每位上数字全不相同而且由奇数构成的整数？解：该题相当于从“1，3，5，7，9”五个数字中分别选出1，2，3，4作排列的方案数；（1）选1个，即构成1位数，共有15P 个；（2）选2个，即构成两位数，共有25P 个；（3）选3个，即构成3位数，共有35P 个；（4）选4个，即构成4位数，共有45P 个；由加法法则可知，所求的整数共有：12345555205P P P P +++=个。

2．比5400小并具有下列性质的正整数有多少个？（1）每位的数字全不同；（2）每位数字不同且不出现数字2与7；解：（1）比5400小且每位数字全不同的正整数；按正整数的位数可分为以下几种情况：① 一位数，可从1～9中任取一个，共有9个；② 两位数。

十位上的数可从1～9中选取，个位数上的数可从其余9个数字中选取，根据乘法法则，共有9981⨯=个；③ 三位数。

百位上的数可从1～9中选取，剩下的两位数可从其余9个数中选2个进行排列，根据乘法法则，共有299648P ⨯=个；④ 四位数。

又可分三种情况：⏹ 千位上的数从1～4中选取，剩下的三位数从剩下的9个数字中选3个进行排列，根据乘法法则，共有3942016P ⨯=个；⏹ 千位上的数取5，百位上的数从1～3中选取，剩下的两位数从剩下的8个数字中选2个进行排列，共有283168P ⨯=个；⏹ 千位上的数取5，百位上的数取0，剩下的两位数从剩下的8个数字中选2个进行排列，共有2856P =个；根据加法法则，满足条件的正整数共有：9816482016168562978+++++=个；（2）比5400小且每位数字不同且不出现数字2与7的正整数；按正整数的位数可分为以下几种情况：设{0,1,3,4,5,6,8,9}A =① 一位数，可从{0}A -中任取一个，共有7个；② 两位数。

十位上的数可从{0}A -中选取，个位数上的数可从A 中其余7个数字中选取，根据乘法法则，共有7749⨯=个；③ 三位数。

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。

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）=35x 1+x 2+x 3+x 4=3 的非负整数解，其个数为F （4，3）=20x 1+x 2+x 3+x 4=2 的非负整数解，其个数为F （4，2）=10x 1+x 2+x 3+x 4=1 的非负整数解，其个数为F （4，1）=4x 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。

第三章3.12.一年级有100名学生参加中文，英语和数学的考试，其中92人通过中文考试，75人通过英语考试，65人通过数学考试；其中65人通过中，英文考试，54人通过中文和数学考试，45人通过英语和数学考试，试求通过3门学科考试的学生数。

[解].令：A 1={通过中文考试的学生}A 2={通过英语考试的学生}A 3={通过数学考试的学生}于是 |Z| =100，|A 1|=92，|A 2|=75，|A 3|=65|A 1∩A 2|=65，|A 1∩A 3|=54，|A 2∩A 3|=45此题没有给出：有多少人通过三门中至少一门；有多少人一门都没通过。

但是由 max{ |A 1|,|A 2|,|A 3| }=max{92,75,65}=92故可以认为：至少有92人通过三门中至少一门考试，即100≥|A 1∪A 2∪A 3|≥92至多有8人没通过一门考试，即0≤|1A ∩2A ∩3A | ≤8于是，根据容斥原理，有|A 1∪A 2∪A 3|=(|A 1|+|A 2|+|A 3|)-(|A 1∩A 2|+|A 1∩A 3|+|A 2∩A 3|)+|A 1∩A 2∩A 3|即 |A 1∩A 2∩A 3|=|A 1∪A 2∪A 3|-(|A 1|+|A 2|+|A 3|)+(|A 1∩A 2|+|A 1∩A 3|+|A 2∩A 3|)=|A 1∪A 2∪A 3|-(92+75+65)+(65+54+45)=|A 1∪A 2∪A 3|-232+164=|A 1∪A 2∪A 3|-68从而由 92-68≤|A 1∪A 2∪A 3|-68≤100-68即 24≤|A 1∪A 2∪A 3|-68≤32可得 24≤|A 1∩A 2∩A 3| ≤32故此，通过3门学科考试的学生数在24到32人之间。

也可用容斥原理，即|1A ∩2A ∩3A |=|Z|-(|A 1|+|A 2|+|A 3|)+(|A 1∩A 2|+|A 1∩A 3|+|A 2∩A 3|)-|A 1∩A 2∩A 3|=100-(92+75+65)+(65+54+45)-|A 1∩A 2∩A 3|=100-232+164-|A 1∩A 2∩A 3|=32－|A 1∩A 2∩A 3|从而有 |A 1∩A 2∩A 3|=32-|1A ∩2A ∩3A |由已知 0≤|1A ∩2A ∩3A |≤8，可得24≤|A 1∩A 2∩A 3|≤32故此，通过3门学科考试的学生数在24到32之间。

第一章答案 第二章答案 第三章答案 第四章答案第一章答案1．(a) 45 ( {1,6},{2,7},{3,8},…,{45,50} )(b) 45⨯5+(4+3+2+1) = 235( 1→2~6, 2→3~7, 3→4~8, …,45→46~50, 46→47~50, 47→48~50, 49→50 ) 2．(a) 5!8!(b) 7! P(8,5) (c) 2 P(5,3) 8! 3. (a) n!P(n+1, m) (b) n!(m+1)!(c) 2!((m+n-2)+1)! 4. 2 P(24,5) 20!5. 2⨯5⨯P(8,2)+3⨯4⨯P(8,2)6. (n+1)!-17. 用数学归纳法易证。

8. 41⨯319. 设 n=p 1n 1p 2n 2…p kn k , 则n 2的除数个数为 ( 2p 1+1) (2p 2+1) …(2p k+1).10．1）用数学归纳法可证n 能表示成题中表达式的形式；2）如果某n 可以表示成题中表达式的形式，则等式两端除以2取余数，可以确定a 1；再对等式两端的商除以3取余数，又可得a 2；对等式两端的商除以4取余数，又可得a 3；…；这说明表达式是唯一的。

11．易用C(m,n)=m!/(n!(m-n)!)验证等式成立。

组合意义：右：从n 个不同元素中任取r+1个出来，再从这r+1个中取一个的全体组合的个数；左：上述组合中，先从n 个不同元素中任取1个出来，每一个相同的组合要生复 C(n-1,r) 次。

12．考虑,)1(,)1(101-=-=+=+=∑∑n nk k k n nnk kknx n x kC x x C 求导数后有令x=1, 即知.210-==∑n nk kn n kC13. 设此n 个不同的数由小到大排列后为a 1, a 2, …, a n 。

当第二组最大数为a k 时，第二组共有2k-1种不同的可能，第一组有2n-k -1种不同的可能。

3.1题（宗传玉）某甲参加一种会议，会上有6位朋友，某甲和其中每人在会上各相遇12次，每二人各相遇6次，每三人各相遇3次，每五人各相遇2次，每六人各相遇一次，1人也没有遇见的有5次，问某甲共参加了几次会议解:设A i为甲与第i个朋友相遇的会议集，i＝1，…，6．则故甲参加的会议数为:28+5=33.3.2题（宗传玉）求从1到500的整数中被3和5整除但不被7整除的数的个数.解:设A3：被3整除的数的集合A5：被5整除的数的集合A7：被7整除的数的集合所以3.3.题（宗传玉）n个代表参加会议，试证其中至少有2人各自的朋友数相等。

解:每个人的朋友数只能取0，1，…，n－1．但若有人的朋友数为0，即此人和其他人都不认识，则其他人的最大取数不超过n－2．故这n个人的朋友数的实际取数只有n－1种可能．，所以至少有２人的朋友数相等．3.4题（宗传玉）试给出下列等式的组合意义.解:(a) 从n 个元素中取k 个元素的组合，总含有指定的m 个元素的组合数为)()(kn mn m k m n --=--。

设这m 个元素为a 1,a 2,…,a m ,Ai 为不含a i 的组合（子集），i=1,…,m.()∑∑∑==∈⊄==⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛-=-+⎪⎪⎭⎫ ⎝⎛==⎪⎪⎭⎫⎝⎛--⎪⎪⎭⎫⎝⎛-=ml l m l l m i i lj i lk l n k m A k n k n m n k l n l j 01),(),...,(1m1i i i i i 1)1(A A A A 111213.5题（宗传玉）设有三个7位的二进制数：a1a2a3a4a5a6a7，b1b2b3b4b5b6b7，c1c2c3c4c5c6c7．试证存在整数i 和j，1≤i≤j≤7，使得下列之一必定成立：a i=a j=b i=b j，a i=a j=c i=c j，b i=b j=c i=c j．证：显然，每列中必有两数字相同，共有种模式，有０或１两种选择．故共有·2种选择．·2=6．现有7列，．即必有２列在相同的两行选择相同的数字，即有一矩形，四角的数字相等．3.6题（宗传玉）在边长为1的正方形内任取5个点试证其中至少有两点，其间距离小于证：把１×１正方形分成四个(1/2)×.则这５点中必有两点落在同一个小正方形内．而小正方形内的任两点的距离都小于．3.7题（王星）在边长为1的等边三角形内任取5个点试证其中至少有两点，期间距离小于1/2．证：把边长为１的三角形分成四个边长为1/2的三角形，如上图：则这５点中必有两点落在同一个小三角形中．小三角形中任意两点间的距离都小于1/2．3.8题（王星）任取11个整数，求证其中至少有两个数它们的差是10的倍数。