组合数学第三章习题解答剖析
组合数学-卢开澄-习题答案
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, 48→49~50, 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. 因首数字可分别为偶数或奇数，知结果为 2⨯5⨯P(8,2)+3⨯4⨯P(8,2).6. (n+1)!-17. 用数学归纳法易证。
8. 两数的公共部分为240530, 故全部公因数均形如2m 5n ,个数为41⨯31. 9. 设有素数因子分解 n=p 1n 11p 2 n 22…p k n k k , 则n 2的除数个数为( 2n 1+1) (2n 2+1) …(2n 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(1010-=-=+=+=∑∑n nk k k n nnk kk nx n x kC x x C 求导数后有令x=1, 即知.210-==∑n nk k n n kC13. 设此n 个不同的数由小到大排列后为a 1, a 2, …, a n 。
卢开澄组合数学--第三章习题解答
9. 题目 由推导过程知 解：
ln( G ( x )) 1 1 2
2
x 1 x 1 3
2
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1 2
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x 1 x
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ln( G ( x )) ln pn n ln ln pn x 1 x x 1 x
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2 n
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A0 A 1 A2 A3
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其中 x 项系数为所求
8
11. 题目 解： 用归纳法可证明： 1）当k=1时命题成立 2）设当k=N时命题成立 即N可唯一表示成不同且不相邻的F 数之和。 则当k=N+1时，明显可以分成N的序列 再加上1（ F2），但这可能会不能满足 “不同且不相邻”的条件。 下面予以讨论
n 1
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6. 题目 解：
参见第四题解答前半部分。
7. 题目 解：题设中序列的母函数为：
G ( x ) C ( n, n) C ( n 1, n) x C ( n k , n) x
13. 题目 解： 设符合条件的n位二进制数的个数为hn 这些数中一共有a n个0 当n位二进制数最高位为1时,符合条件 的n位二进制数的个数为 hn 1
组合数学(第3章3.3)
多重集的排列及组合
主要内容
多重集排列应用 多重集的组合及应用
回顾：多重集排列计数
定理3.4.2：令S是多重集，它有k个不同的 元素，每个元素的重复数分别为n1，n2，…， nk，那么，S的排列数等于 n! n1! n 2 !… n k ! 其中n= n1+n2+…+nk
多重集排列与集合划分
6 = 84
解：（1）方程x1+x2+x3+x4=10 （B）的正整数
例
3. 方程x1+x2+x3+x4=20的整数解的个数是多少？其中 x1≥3, x2≥1, x3≥0, x4≥5. 解：作变量代换：y1=x13， y2=x21， y3=x3， y4=x45，那么，得到方程： y1+y2+y3+y4=11。原 方程的解个数与该方程的非负整数解个数相同。 故为：
r + k - 1 r + k - 1 = r k -1
定理的证明
(1) 令S＝{∞a1, ∞a2,…, ∞ak}，那么S的一个r-组合 具有形式{x1a1, x2a2,…, xkak}，其中 x1+x2+…+xk=r （A） A xi是非负整数。 (2) 方程（A）的任何一个解确定S的一个r-组合，因 此，S的r-组合个数等于方程（A）解的个数。
11 + 4 1 14 11 = 11
问题？
令多重集S={n1a1, n2a2, …, nkak}，n= n1+n2+…+nk ，求S的r-组合数，其中0≤r≤n. 方程： x1+x2+…+xk=r 满足条件 0≤x1≤n1，0≤x2≤n2，…, 0≤xk≤nk 的整数解的个数。
组合数学_第3章3.1_ (1)
显然，当k1 k2, 则n2 整除n1，否则n1整除n2。
例：对于任意给定的52个非负整数，证明：其中必存 在两个非负整数，要么两者的和能被100整除，要么 两者的差能被100整除。
证：对于任意一个非负整数，其整除100的余数可能 为{0, 1, 2, …, 99}中之一。 对这100个余数进行分组，构造如下51个集合：
何整数n, n = 2k a ，其中，a为奇数，k0。
❑ 200内只能有100个不同奇数，故可对101个 数运用鸽巢原理。
例. 从整数1, 2, ，200中选取101个整数。证明 所选的数中存在两个整数，使得其中一个是另 一个的因子。
证：对于1到200间的整数n，n可写作以下形 式：n = 2k a , 其中a只能是200内的奇数。 由于要选取101个整数，但是 200内只有100个奇 数，应用鸽巢原理知必存在两个数n1与n2除以2 的余数相等。假设
思考：随意地把一个3行9列棋盘的每个方格涂成红色 或蓝色，求证：必有两列方格的涂色方式是一样的。
1 23 456 78 9
每列的涂色方式一共有23= 8种
思考：
（英国数学奥林匹克1975年的问题）在一个半 径为1单位的圆板上钉7个钉，使得两个钉的 距离是大于或等于1，那么这7个钉一定会有一 个位置恰好是在圆心上。
吾尝从君济于河，鼋 衔左骖以入砥柱之流。 当是时也，冶少不能
吾仗兵而却三军者再，若开疆之功， 游，潜行逆流百步，
亦可以食桃，而无与人同矣
顺流九里，得鼋而杀
之，左操骖尾，右挈
鼋头，鹤跃而出
◼ 宋代费衮的《梁溪漫志》中，就曾运用抽屉原理 来批驳“算命”一类迷信活动的谬论
《组合数学》教案 3章(递推关系)及课后习题讲解
第三章 递推关系§3.1 基本概念（一） 递推关系【定义3.1.1】（隐式）对数列{}0≥i a i 和任意自然数n ，一个关系到n a 和某些个i a （i ＜n ）的方程式，称为递推关系，记作()0,,,10=n a a a F例 022022212=-------n a a a a n n n 01223121=-------a a a a n n n 【定义3.1. 1'】（显式） 对数列{}0≥i a i ，把n a 与其之前若干项联系起来的等式对所有n ≥k 均成立（k 为某个给定的自然数），称该等式为{}i a 的递推关系，记为()k n n n n a a a F a ---=,,,21 （3.1.1）＇例 1223121++++=--a a a a n n n（二） 分类（1）按常量部分：① 齐次递推关系：指常量＝0，如21--+=n n n F F F ； ② 非齐次递推关系，即常量≠0，如121=--n n h h 。
（2）按i a 的运算关系：① 线性关系，F 是关于i a 的线性函数，如（1）中的nF 与n h 均是如此；② 非线性关系，F 是i a 的非线性函数，如112211h h h h h h h n n n n ---+++= 。
（3）按i a 的系数：① 常系数递推关系，如（1）中的n F 与n h ；② 变系数递推关系，如1-=n n np p ，1-n p 之前的系数是随着n 而变的。
（4） 按数列的多少：① 一元递推关系，其中的方程只涉及一个数列，如（3.1.1）和（3.1.1）＇均为一元的；② 多元递推关系，方程中涉及多个数列，如⎩⎨⎧+=+=----111177n n n n n n a b b b a a （5）显式与隐式：⎥⎦⎤⎢⎣⎡-+=++++11112n n n n n y x y h y y （三） 定解问题 【定义3.1.2】（定解问题）称含有初始条件的递推关系为定解问题，其一般形式为()⎩⎨⎧====--11110010,,,,0,,,k k n d a d a d a a a a F （3.1.2）解递推关系：求n a 的与a 0、a 1、…、a n －1无关的解析表达式或数列{}n a 的母函数。
组合数学第三章课后习题答案
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。
李凡长版组合数学课后习题答案习题3
f(2)=1,f(3)=1,f(4)=2.
6. 求 n 位 0,1 序列中“ 010”只出现一次且在第 n 位出现的序列数 f(n). 解：最后三位是“ 010”的序列共有 2n-3 个。包括以下情况：
f(n) 包含了在最后三位第一次出现 010 的个数，同时排除了从
n-4 到 n-2 位第一次出现 010 的可能；
13. 在一个平面上画一个圆 , 然后一条一条地画 n 条与圆相交的直线 . 当 r 是 大于 1 的奇数时 , 第 r 条直线只与前 r －1 条直线之一在圆内相交 . 当 r 是偶数时 , 第 r 条直线与前 r －1 条直线都在圆内相交 . 如果无 3 条直线在 圆内共点 , 这 n 条直线把圆分割成多少个不重叠的部分？
2) 证明 f r (n, k)
n rk r , n r k( r 1)
k
解：可将本题转换为构造相应的 0－1 串的问题。将这样的 n 位 0－1 串与 1 到 n 的正整数对位，与 1 相应的整数选取，与 0 相应的不取。一个 0－1 串 对应一个选取方案。这也对应将相同的球放入不同的盒子的方案数。
解：设 f(n) 表示 n 个椭圆将平面分割成的部分的个数， 则有： 一个椭圆将平
面分成内、外两个部分，两个椭圆将平面分成 4 个部分。第二个椭圆的周界
被第一个椭圆分成两部分，这恰恰是新增加的域的边界。依此类推，第三个
椭圆曲线被前面两个椭圆分割成 4 部分，将平面分割成 4＋4＝8 个部分。若
n－ 1 个椭圆将平面分割成 f(n-1) 个部分， 第 n 个椭圆和前 n－1 个椭圆两两
f (n) (n 2) f (n 1) ( n 1)
（6）
；
f (0) 1
解： f(n)=(n+2)f(n-1)=(n+2)(n+1)f(n-2)=
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。
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3.5 设有3个7位的二进制数 a1 a2 a3 a4 a5 a6 a7, b1 b2 b3 b4 b5 b6 b7, c1 c2 c3 c4 c5 c6 c7,
试证存在整数i和j，1≤i<j≤7，使得下列之一必然成立： ai=aj=bi=bj,ai=aj=ci=cj,bi=bj=ci=cj
证明：
i 1
Ai 12C(6,1) 6C(6,2) 4C(6,3) 3C(6,4) 2C(6,5) C(6,6) 72 90 80 45 12 1 28
28 5 33
共33次会议
2.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
A7 A5 A3 A5 A3 A7 A5 A3
500 3 5
3
500 5
7
33
4
29
3. n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．以下分两种 情况讨论。
若有人的朋友数为0，即此人和其他人都不认识，则其他n1个人的最大取数不超过n－2．必有两人认识人数相等。 若没有人的朋友数为0，则这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。
A1 A2 ... Ak C(n m l 1,l) C(n m,1)C(n m l 2,l)
C(n m,2)C(n m l 3,l) ... (1)nm C(n m, n m)C(l 1,l)
n-m
(1) j C(n m, j)(n m j l 1,l)
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
(1)2 (1)2 2 22 2
3.7 在边长为1的等边三角形内任取5点，试证至少有两点距离 小于1/2
证把边长为１的三角形分成四个边长为－的三角形，如下图： 则这５点中必有两点落在同一个小三角 形中．
3.8 任取11个数，求证其中至少有两个数它们的差是10的倍数。
Ai1 Ai2 ... Aim C(n, k) C(m,1)C(n 1, k) C(m,2)C(n 2, k) ...
(1)m C(m, m)C(n m, k)
m
(1)l C(m,l)(n l, k)
i1
( )令k＝n－m．l个相同的球放入k个不同的盒子里．每盒不 空的方案数为C(n-m+l-n+m-1,l-n+m)=C(l-1,n-m-1)。设Ai 为第i个盒子为空的方案集，i=1,2,…,k.
证明：一个数是不是10的倍数取决于这个数的个位数是不是0， 是 0就是10的倍数。 一个数的个位数只可能是0,1,...,9十个数，任取11个数，其中必 有两个个位数相同，那么这两个数的差的个位数必然是0。
3.9 把从1到326的326个整数任意分为5个部分，试证其中有一部 分至少有一个数是某两个数之和，或是另一个数的两倍。 证明：用反证法，设存在划分。
第三章习题
3.1 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4次，每四人 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？
1．解 设Ai为甲与第i个朋友相遇的会议集，i＝1，…，6．则
6
Ai 12 C(6,1)
(a) C(n m, n k) C(n m, k m) (a)从n个元素中取k个元素的组合，总含m个指定的元素的组合 数为C(n-m,k-m),设这m个元素为a1,a2,...,am.Ai为不含ai的组 合，i=1,2,...,m
Ai C(n 1, k) Ai1 Ai2 ... Ail C(n l, k)
3.4 试给下列等式以组合意义
m
(a) C(n m, n k) (1)l C(m,l)C(n l, k), n k m i0 nm
(b) C(l 1, n m 1) (1) j C(n m, j)C(n m j l 1,l) j0
(c) C(m l 1, m 1) C(m l, m) C(m l, m 1) C(m l, m 2) ... (1)l C(m l, m l)
显然，每列中必有两数字相同，共有C(3,2)种模式，有０或１两 种选择．故共有C(3,2)·2种选择。C(3,2)·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 在边长为1的正方形内任取5点，试证其中至少有两点，其间 距离小于 2 证 把１×2１正方形分成四个相等的小正 方形．如下图：
j1
l个相同的球放入n个不同的盒子里，指定的m个盒子为空，其他 盒子不空的方案数为C(l-1,n-m-1)
( c ) 设Ai为m+l个元素中取m+i个，含特定元素a的方案集；Ni 为m+l个元素中取m+i个的方案数．则：
Ni C(m l, m i) Ai C(m l 1, m i 1) Ai Ni Ai C(m l, m i) C(m l 1, m i 1)
C(m l 1, m i) Ai1
Ai Ni Ai ,i 1,2,...,l
A0 N0 A0 N0 A1 N0 (N1 A1 ) ... N0 N1 N2 ... (1)l Nl C(m l, m) C(m l, m 1) C(m l, m 2) ... C(m l, m l)
P1 P2 P3 P4 P5 {1,...,326}
因为32651 1 66,因此必存在有一子集含66个元素 不妨设为P1
设这66个元素为a1<a2<a3<...<a66
构造b1=a2-a1, b2=a3-a1,…, b65=a66-a1,
令B={b1,b2,…,b65}
这65个元素属于1到326，如果这65个元素有任何一个属于P1， 则定理得证。