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2016-2017《创新设计》同步人教A版选修2-3第一章 1.1(二)

2016-2017《创新设计》同步人教A版选修2-3第一章 1.1(二)
2016-2017《创新设计》同步人教A版选修2-3第一章 1.1(二)

[学习目标] 1.进一步理解和掌握分类加法计数原理和分步乘法计数原理.2.能根据实际问题特征,正确选择计数原理解决实际问题.

知识点两个计数原理的联系与区别

题型一两个计数原理在排数中的应用

例1用0,1,2,3,4,5可以组成多少个无重复数字且比2 000大的四位偶数?

解完成这件事可分为三类:

第一类是个位数字为0的比2 000大的四位偶数,可以分三步完成:

第一步,选取千位上的数字,只有2,3,4,5可以选择,有4种选法;

第二步,选取百位上的数字,除0和千位上已选定的数字以外,还有4个数字可以选择,有4种选法;

第三步,选取十位上的数字,有3种选法.

由分步乘法计数原理知,这类数的个数为4×4×3=48.

第二类是个位数字为2的比2 000大的四位偶数,可以分三步完成:

第一步,选取千位上的数字,除去2,1,0只有3个数字可以选择,有3种选法;

第二步,选取百位上的数字,在去掉已经确定的首尾2个数字之后,还有4个数字可以选择,有4种选法;

第三步,选取十位上的数字,有3种选法.

由分步乘法计数原理知,这类数的个数为3×4×3=36.

第三类是个位数字为4的比2 000大的四位偶数,其方法步骤同第二类.

对以上三类用分类加法计数原理,得所求无重复数字且比2 000大的四位偶数有48+36+36=120(个).

反思与感悟排数问题实际就是分步问题,需要用分步乘法计数原理解决.此题中,由于数字0的出现,又进行了分类讨论,即在解决相关的排数问题时,要注意两个原理的综合应用.跟踪训练1用0,1,2,3,4这5个数字可以组成多少个按下列要求的无重复数字?

(1)四位密码;

(2)四位数;

(3)四位奇数.

解(1)完成“组成无重复数字的四位密码”这件事,分为四个步骤:

第一步,取左边第一位上的数字,有5种选取方法;

第二步,取左边第二位上的数字,有4种选取方法;

第三步,取左边第三位上的数字,有3种选取方法;

第四步,取左边第四位上的数字,有2种选取方法.

由分步乘法计数原理知,可以组成不同的四位密码共有

N=5×4×3×2=120(个).

(2)方法一完成“组成无重复数字的四位数”这件事分为四个步骤:

第一步,从1,2,3,4中选取一个数字作千位数字,有4种选取方法;

第二步、第三步、第四步与(1)类似,分别有4,3,2种选取方法.

由分步乘法计数原理知,可以组成不同的四位数共有

N=4×4×3×2=96(个).

方法二与第(1)问的区别在于:四位密码首位可以是0,而四位数首位不可以为0.因此,只需求首位为0的四位密码有多少个,用(1)的总数减去首位为0的个数即为所求.

当首位是0时,第二位有4种选取方法,第三位有3种选取方法,第四位有2种选取方法,由分步乘法计数原理知,首位是0的四位密码共有1×4×3×2=24(个).

故无重复数字的四位数共有120-24=96(个).

(3)完成“组成无重复数字的四位奇数”这件事,分两类方案.

第一类:这个四位奇数的个位数字是1,分三个步骤去完成.

第一步,选取千位上的数字,有3种(从2,3,4中选)不同选法;

第二步,选取百位上的数字,有3种不同选法;

第三步,选取十位上的数字,有2种不同选法.

由分步乘法计数原理知,该类中四位奇数共有1×3×3×2=18(个).

第二类:这个四位奇数的个位数字是3,也是分三个步骤去完成.

具体求法与个位数字是1时完全一样,因而这样的奇数也是18个.

由分类加法计数原理知,共可组成无重复数字的四位奇数18+18=36(个).

题型二抽取(分配)问题

例2现有3名医生、5名护士、2名麻醉师.

(1)从中选派1名去参加外出学习,有多少种不同的选法?

(2)从这些人中选出1名医生、1名护士和1名麻醉师组成1个医疗小组,有多少种不同的选法?

解(1)分三类:

第一类,选出的是医生,有3种选法;

第二类,选出的是护士,有5种选法;

第三类,选出的是麻醉师,有2种选法.

根据分类加法计数原理,共有3+5+2=10(种)选法.

(2)分三步:

第一步,选1名医生,有3种选法;

第二步,选1名护士,有5种选法;

第三步,选1名麻醉师,有2种选法.

根据分步乘法计数原理知,共有3×5×2=30(种)选法.

反思与感悟解决抽取(分配)问题的方法

(1)当涉及对象数目不大时,一般选用例举法、树状图法、框图法或者图表法.

(2)当涉及对象数目很大时,一般有两种方法:①直接使用分类加法计数原理或分步乘法计数原理.一般地,若抽取是有顺序的就按分步进行;若是按对象特征抽取的,则按分类进行.②间接法:去掉限制条件,计算所有的抽取方法数,然后减去所有不符合条件的抽取方法数即可.

跟踪训练23个不同的小球放入5个不同的盒子,每个盒子至多放一个小球,共有多少种方法?

解方法一(以小球为研究对象)分三步来完成:

第一步,放第一个小球有5种选择;

第二步,放第二个小球有4种选择;

第三步,放第三个小球有3种选择.

根据分步乘法计数原理得,共有方法数N=5×4×3=60(种).

方法二(以盒子为研究对象)盒子标上序号1,2,3,4,5;分成以下10类:

第一类,空盒子标号为(1,2),选法有3×2×1=6(种);

第二类,空盒子标号为(1,3),选法有3×2×1=6(种);

第三类,空盒子标号为(1,4),选法有3×2×1=6(种).

分类还有以下几种情况:(1,5),(2,3),(2,4),(2,5),(3,4),(3,5),(4,5),共10类,每一类都有6种方法.

根据分类加法计数原理得,共有方法数N=6+6+…+6=60(种).

题型三涂色问题

例3将红、黄、绿、黑4种不同的颜色分别涂入如图所示的5个区域内,要求相邻的两个区域的颜色不相同,问:有多少种不同的涂色方法?

解将5个区域分别标记为A,B,C,D,E(如图所示),则A区域有4种不同的涂色方法,B区域有3种,C区域有2种,D区域有2种,但E区域的涂色依赖于B区域与D区域涂的颜色,如果B区域与D区域涂的颜色相同,则E区域有2种涂色方法,如果B区域与D区域所涂的颜色不相同,则E区域只有1种涂色方法.因此应先分类后分步.

(1)当B与D同色时,有4×3×2×2=48(种)涂色方法;

(2)当B与D不同色时,有4×3×2×1×1=24(种)涂色方法.

故共有48+24=72(种)不同的涂色方法.

反思与感悟(1)涂色问题的基本要求是相邻区域不同色,但是不相邻的区域可以同色.解决此类问题要特别关注图形的结构特征.如果图形不很规则,往往从某一块出发进行分步涂色,从而选用分步乘法计数原理;如果图形具有一定的对称性,那么先对涂色方案进行分类,每一类再进行分步.

(2)涂色问题往往涉及两计数原理的综合应用,因此,要找准分类标准,兼顾条件的情况下分步涂色.

跟踪训练3如图所示,用5种不同的颜料给4块图形(A,B,C,D)涂色,要求共边两块颜色互异,求有多少种不同的涂色方案.

解方法一按A,C颜色相同或不同进行分类.

若A,C颜色相同,则A有5种涂色方法,B有4种涂色方法,D有4种涂色方法,故共有5×4×4=80(种)涂法.

若A,C颜色不同,则A有5种涂色方法,C有4种涂色方法,B有3种涂色方法,D有3种涂色方法,故共有5×4×3×3=180(种)涂法.

根据分类加法计数原理,共有80+180=260(种)不同的涂色方案.

方法二按涂色种类进行分类.

第一类:涂4种颜色,分四步,A有5种涂法,B有4种涂法,C有3种涂法,D有2种涂法.

故共有5×4×3×2=120(种)涂法.

第二类:涂3种颜色,则A,C颜色相同或B,D颜色相同.

当A,C颜色相同时,A,C有5种涂法,B有4种涂法,D有3种涂法.

故共有5×4×3=60(种)涂法.

当B,D颜色相同时,同理也有60种不同的涂法.

故共有60+60=120(种)涂法.

第三类:涂2种颜色,则A,C颜色相同,B,D颜色相同,A,C有5种涂法,B,D有4种涂法.

故共有5×4=20(种)涂法.

根据分类加法计数原理,共有120+120+20=260(种)不同的涂色方案.

题型四种植问题

例4从黄瓜、白菜、油菜、扁豆4种蔬菜品种中选出3种,分别种在不同土质的三块土地上,其中黄瓜必须种植,求有多少种不同的种植方法.

解方法一(直接法):若黄瓜种在第一块土地上,则有3×2×1=6(种)不同种植方法.

同理,黄瓜种在第二块、第三块土地上,均有3×2×1=6(种)不同种植方法.

故不同的种植方法共有6×3=18(种).

方法二(间接法):从4种蔬菜中选出3种,种在三块地上,有4×3×2=24(种),其中不种黄瓜有3×2×1=6(种),故共有不同种植方法24-6=18(种).

反思与感悟按元素性质分类,按事件发生过程分步是计数问题的基本思想方法,区分“分类”与“分步”的关键,是验证所提供的某一种方法是否完成了这件事情,分类中的每一种方法都完成了这件事情,而分步中的每一种方法不能完成这件事情,只是向事情的完成迈进了一步.

跟踪训练4一个同心圆形花坛,分为两部分,中间小圆部分种植草坪和绿色灌木,周围的圆环分为n(n≥3,n∈N)等份,种植红、黄、蓝三色不同的花,要求相邻两部分种植不同颜色的花.

(1)如图1,圆环分成的3等份为a1,a2,a3,有多少种不同的种植方法?

(2)如图2,圆环分成的4等份为a1,a2,a3,a4,有多少种不同的种植方法?

解(1)如图1,先对a1部分种植,有3种不同的种植方法,再对a2,a3种植.

因为a2,a3与a1不同颜色,a2,a3也不同,所以由分步乘法计数原理得3×2×1=6(种).(2)如图2,当a1,a3不同色时,有3×2×1×1=6(种)种植方法,当a1,a3同色时,有3×2×2=12(种)种植方法,由分类加法计数原理,共有6+12=18(种)种植方法.

用分类讨论思想解决两类计数问题

例5用0,1,2,3,…,9十个数字可能组成多少个不同的

(1)三位数;

(2)小于500且没有重复数字的自然数.

分析由于0不能在首位,所以解决本题可以从特殊元素0入手讨论,也可从特殊位置(最高位)入手讨论.

解(1)由于0不能在首位,所以首位数字有9种选法,十位与个位上的数字均有10种选法,所以不同的三位数共有9×10×10=900(个).

(2)一位自然数有10个,二位自然数有9×9=81(个),三位自然数有4×9×8=288(个).

所以共有10+81+288=379(个)小于500且无重复数字的自然数.

点评解决计数问题时要善于挖掘题目中的限制条件.要注意特殊优先的原则,即优先处理特殊位置、特殊元素,如果从正面不好解决,可采用间接法求解.

1.已知函数y=ax2+bx+c为二次函数,其中a,b,c∈{0,1,2,3,4},则不同的二次函数的个数为()

A.125 B.15

C.100 D.10

答案 C

解析若y=ax2+bx+c为二次函数,则a≠0,要完成该事件,需分步进行:

第一步,对于系数a有4种不同的选法;

第二步,对于系数b有5种不同的选法;

第三步,对于系数c有5种不同的选法.

由分步乘法计数原理知,共有4×5×5=100(个).

2.6把椅子摆成一排,3人随机就座,任何两人不相邻的坐法种数为()

A.144 B.120

C.72 D.24

答案 D

解析剩余的3个座位共有4个空隙供3人(不妨记为甲、乙、丙)选择就座,因此,可分三步:甲从4个空隙中任选一个空隙,有4种不同的选择;乙从余下的3个空隙中任选一个空隙,有3种不同的选择;丙从余下的2个空隙中任选一个空隙,有2种不同的选择.根据分步计数原理,任何两人不相邻的坐法种数为4×3×2=24.故选D.

3.两人进行乒乓球比赛,采取五局三胜制,即先赢三局者获胜,决出胜负为止,则所有可能出现的情形(各人输赢局次的不同视为不同情形)共有()

A.10种B.15种

C.20种D.30种

答案 C

解析由题意知,比赛局数最少为3局,至多为5局.当比赛局数为3局时,情形为甲或乙连赢3局,共2种;当比赛局数为4局时,若甲赢,则前3局中甲赢2局,最后一局甲赢,共有3种情形;同理,若乙赢,则也有3种情形,所以共有6种情形;当比赛局数为5局时,前4局,甲、乙双方各赢2局,最后一局胜出的人赢,若甲前4局赢2局,共有赢取第1、2局,1、3局,1、4局,2、3局,2、4局,3、4局六种情形,所以比赛局数为5局时共有2×6=12(种),综上可知,共有2+6+12=20(种).故选C.

4.(a1+a2)·(b1+b2+b3)·(c1+c2+c3+c4)的展开式中有________项.

答案24

解析要得到项数分三步:

第一步,从第一个因式中取一个因子,有2种取法;

第二步,从第二个因式中取一个因子,有3种取法;

第三步,从第三个因式中取一个因子,有4种取法.

由分步乘法计数原理知,共有2×3×4=24(项).

5.将3种作物种植在如图所示的5块试验田里,每块种植一种作物且相邻的试验田不能种植同一种作物,不同的种植方法共有________种(以数字作答).

答案42

解析分别用a,b,c代表3种作物,先安排第一块田,有3种方法,不妨设放入a,再安排第二块田,有2种方法b或c,不妨设放入b,第三块也有2种方法a或c.

(1)若第三块田放c:

第四、五块田分别有2种方法,共有

(2)若第三块田放a:

第四块有b或c2种方法:

①若第四块放c:

第五块有2种方法;

②若第四块放b:

第五块只能种作物c,共1

综上,共有3×2×(2×2+2+1)=42(种)方法.

1.分类加法计数原理与分步乘法计数原理是两个最基本,也是最重要的原理,是解答排列、组合问题,尤其是较复杂的排列、组合问题的基础.

2.应用分类加法计数原理要求分类的每一种方法都能把事件独立完成;应用分步乘法计数原理要求各步均是完成事件必须经过的若干彼此独立的步骤.

3.一般是先分类再分步,分类时要设计好标准,设计好分类方案,防止重复和遗漏.4.若正面分类,种类比较多,而问题的反面种类比较少时,则使用间接法会简单一些.

一、选择题

1.已知集合S={a1,a2},T={b1,b2},则从集合S到T的映射共有()

A.1个B.2个

C.3个D.4个

答案 D

解析可分两步,第一步,集合S中a1对应到集合T中的元素有2种不同的选法;第二步,集合S中a2对应到集合T中的元素有2种不同的选法.由分步乘法计数原理知,从集合S到T的映射共有2×2=4(个).故选D.

2.从1,2,3,…,10这10个数中任意选出3个不同的数,使这3个数成等比数列,这样的等

比数列的个数为( )

A .3

B .4

C .6

D .8

答案 D

解析 当公比为2时,等比数列可为1,2,4或2,4,8.

当公比为3时,等比数列可为1,3,9.

当公比为32

时,等比数列可为4,6,9. 同时,4,2,1,;8,4,2;9,3,1和9,6,4也都是等比数列,共8个.

3.已知x ∈{1,2,3,4},y ∈{5,6,7,8},则xy 可表示不同值的个数为( )

A .2

B .4

C .8

D .15

答案 D

解析 完成xy 这件事分两步:

第一步:从集合{1,2,3,4}选一个数,共有4种选法;

第二步:从集合{5,6,7,8}选一个数,共有4种选法.

共有4×4=16(种)选法.其中3×8=4×6,所以xy 可表示的不同值的个数为15.

4.有4位教师在同一年级的4个班中各教1个班的数学,在数学检测时要求每位教师不能在本班监考,则监考的方法有( )

A .8种

B .9种

C .10种

D .11种

答案 B

解析 设4位监考教师分别为A ,B ,C ,D,4个班级分别为a ,b ,c ,d ,假设A 监考b ,则余下3人监考剩下的3个班,共有3种不同方法.同理A 监考c 或d 时,也分别有3种不同方法.根据分类加法计数原理,监考的方法共有3+3+3=9(种).

5.满足a ,b ∈{-1,0,1,2},且关于x 的方程ax 2+2x +b =0有实数解的有序数对(a ,b )的个数为( )

A .14

B .13

C .12

D .10

答案 B

解析 ①当a =0时,方程表示垂直于x 轴的直线方程,有解,此时b 取4个值,故有4种有序数对;②当a ≠0时,需要Δ=4-4ab ≥0,即ab ≤1,有3个实数对不满足题意,分别为(1,2),(2,1),(2,2).∵(a ,b )共有3×4=12(个)实数对,此时(a ,b )的取值为12-3=9(个).∴(a ,b )的个数为4+9=13.

6.如图所示,“中国印”被中间的白色图案分成了5个区域,现给它着色,要求相邻区域不能用同一颜色,如果只有4种颜色可供使用,那么不同的着色方法有( )

A.120种B.72种

C.48种D.24种

答案 B

解析以所选颜色的种数为标准,可分两类进行:第一类,用3种颜色有4×3×2=24(种);第二类,用4种颜色有4×3×2×2=48(种),∴共有24+48=72(种)不同方法.故选B. 7.方程ay=b2x2+c中的a,b,c∈{-3,-2,0,1,2,3},且a,b,c互不相同.在所有这些方程所表示的曲线中,不同的抛物线共有()

A.60条B.62条

C.71条D.80条

答案 B

解析利用两个计数原理结合分类讨论思想求解.

当a=1时:若c=0,则b2有2个取值,共2条抛物线;若c≠0,则c有4个取值,b2有2个取值,共有2×4=8(条)抛物线.

当a=2时:若c=0,则b2有3个取值,共有3条抛物线;若c≠0,当c取1时,b2有2个取值,共有2条抛物线,当c取-2时,b2有2个取值,共有2条抛物线,当c取3时,b2有3个取值,共有3条抛物线,当c取-3时,b2有3个取值,共有3条抛物线.

∴a=2时共有3+2+2+3+3=13(条)抛物线.

同理,a=-2,-3,3时,共有抛物线3×13=39(条).

由分类加法计数原理知,共有抛物线39+13+8+2=62(条).

二、填空题

8.五个工程队承建某项工程的5个不同的子项目,每个工程队承建1项,其中甲工程队不能承建1号子项目,则不同的承建方案有________种.

答案96

解析完成承建任务可分五步:

第一步,安排1号有4种;

第二步,安排2号有4种;

第三步,安排3号有3种;

第四步,安排4号有2种;

第五步,安排5号有1种.

由分步乘法计数原理知,共有4×4×3×2×1=96(种).

9.有10本不同的数学书,9本不同的语文书,8本不同的英语书,从中任取两本不同类的书,共有________种不同的取法.

答案 242

解析 分三类:第一类,取数学书和语文书,有10×9=90(种);第二类,取数学书和英语书,有10×8=80(种);第三类,取语文书和英语书,有9×8=72(种).故共有90+80+72=242(种).

10.在2015年田径挑战赛上,8名男运动员参加100米决赛,其中甲、乙、丙3人必须在1,2,3,4,5,6,7,8八条跑道的奇数号跑道上,则安排这8名运动员比赛的方式共有________种. 答案 2 880

解析 分两步安排这8名运动员,第一步,安排甲、乙、丙3名运动员,共有1,3,5,7四条跑道可安排,所以安排方式有4×3×2=24(种);第二步,安排另外5名运动员,可在2,4,6,8及余下的一条奇数号跑道安排,所以安排方式有5×4×3×2×1=120(种).所以安排这8名运动员比赛的方式有24×120=2 880(种).

11.现有某类病毒记作X m Y n ,其中正整数m ,n (m ≤7,n ≤9)可以任意选取,则m ,n 都取到奇数的概率为________.

答案 2063

解析 因为正整数m ,n 满足m ≤7,n ≤9,所以(m ,n )所有可能的取值有7×9=63(种),其

中m ,n 都取到奇数的情况有4×5=20(种),因此所求概率为2063

. 三、解答题

12.如图所示,将四棱锥S -ABCD 的每一个顶点染上一种颜色,并使同一条棱上的两端点异色,如果只有5种颜色可供使用,求不同的染色方法总数.

解 方法一 由题意,四棱锥S -ABCD 的顶点S ,A ,B 所染的颜色互不相同,它们共有5×4×3=60(种)染色方法.当S ,A ,B 染色确定时,不妨设其颜色分别为1,2,3.若C 染2,则D 可染3或4或5,有3种染法;若C 染4,则D 可染3或5,有2种染法;若C 染5,则D 可染3或4,有2种染法.

由分类加法计数原理,当S ,A ,B 已染确定时,C ,D 有7种染法.

由分步乘法计数原理得,不同的染色方法有60×7=420(种).

方法二 第一步,S 点染色,有5种方法.第二步,A 点染色,由于A 与S 在同一条棱上,所以有4种方法.第三步,B 点染色,由于B 与S ,A 分别在同一条棱上,所以有3种方法.第四步,C 点染色,也有3种方法,但考虑到D 点与S ,A ,C 相邻,需要针对A 与C 是否同

色进行分类.

当A 与C 同色时,D 点有3种染色方法,由分步乘法计数原理,有5×4×3×1×3=180(种)方法;当A 与C 不同色时,因为C 与S ,B 也不同色,所以C 点有2种染色方法,D 点也有2种染色方法,再由分步乘法计数原理,有5×4×3×2×2=240(种)方法.由分类加法计数原理得,不同的染色方法共有180+240=420(种).

方法三 第一类,5种颜色全用,有5×4×3×2×1=120(种)不同的染色方法;

第二类,只有4种颜色,则必有某两个顶点同色(A 与C 或B 与D ),共有5×4×3×2+5×4×3×2=240(种)不同的染色方法;第三类,只用3种颜色,则A 与C 、B 与D 必定同色,有5×4×3=60(种)不同的染色方法.

由分类加法计数原理得,不同的染色方法共有120+240+60=420(种).

13.某中学调查了某班全部45名同学参加书法社团和演讲社团的情况,数据如下表:(单位:人)

(1)(2)在既参加书法社团又参加演讲社团的8名同学中,有5名男同学A 1,A 2,A 3,A 4,A 5,3名女同学B 1,B 2,B 3.现从这5名男同学和3名女同学中各随机选出1人,求A 1被选中且B 1未被选中的概率.

解 (1)由调查数据,知既未参加书法社团又未参加演讲社团的有30人,故至少参加上述一个社团的共有45-30=15(人),所以从该班随机选1名同学,该同学至少参加上述一个社团

的概率P =1545=13

. (2)从这5名男同学和3名女同学中各随机选出1人,其所有可能的结果有5×3=15(种). 根据题意,知这些基本事件的出现是等可能的.

事件“A 1被选中且B 1未被选中”所包含的基本事件有{A 1,B 2},{A 1,B 3},共2个.

因此A 1被选中且B 1未被选中的概率P =215.

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