2012年金版新学案新编高三总复习第十一章 第2课时

(本栏目内容，在学生用书中以活页形式分册装订！)

一、选择题

1．已知集合A＝{1,2,3,4}，B＝{5,6,7}，C＝{8,9}．现在从这三个集合中取出两个集合，再从这两个集合中各取出一个元素，组成一个含有两个元素的集合，则一共可以组成的集合个数为()

A．24 B．36

C．26 D．27

解析：分三类：

第一类：选集合A、B可组成C14C13＝12个集合；

第二类：选集合A、C可组成C14C12＝8个集合；

第三类：选集合B、C可组成C13C12＝6个集合．

由分类加法计数原理，可组成12＋8＋6＝26个集合．

答案： C

2．某会议室第一排共有8个座位，现有3人就座，若要求每人左右均有空位，那么不同的坐法种数为()

A．12 B．16

C．24 D．32

解析：插空法，两端的不能插．

○×○×○×○×○共有4个空．故有A34＝24种方法．

答案： C

3．编号为1、2、3、4、5的5个人分别去坐编号为1、2、3、4、5的五个座位，其中有且只有两个人的编号与座位号一致的坐法有()

A．10种B．20种

C．30种D．60种

解析：五个人有两个人的编号与座位号相同，此两人的选法共有C25，假如编号1、2号人坐的号为1、2，其余三人的编号与座号不同，共有2种坐法．

∴符合题意的坐法有2×C25＝2×10＝20种．

答案： B

4．四张卡片上分别标有数字“2”“0”“0”“9”，其中“9”可当“6”用，则由这四张卡片可组成不同的四位数的个数为()

A．6 B．12

C．18 D．24

解析：特殊元素优先处理，先在后三位中选两个位置填两个数字“0”“0”有C23种填法，再决定用“9”还是“6”有两种可能，最后排另两个卡片有A22种排法，所以共可排成C23·2·A22＝12个四位数，故选B.

答案： B

5．有6名男同学和4名女同学自左至右站成一排，其中女同学不相邻而且最右端必须是女同学的排法有________种()

A．A66A44B．C14A36A66

C．C14C36C66D．A66A36

解析：先从4个女生中取一人站在最右端有C14种方法，把六个男生进行全排列，将3个女生插入6个男生的六个空中，有A66·A36种，共有C14A36A66种排法．

答案： B

6．有4个标号为1,2,3,4的红球和4个标号为1,2,3,4的白球，从这8个球中任取4个球排成一排，若取出的4个球的数字之和为10，则不同的排法种数是() A．384 B．396

C．432 D．480

解析：由数字含4的个数由多到少分类：①所取标号为4,4,1,1，共有排法N1＝A44＝24(种)；②所取标号为4,3,2,1，共有排法N2＝24·A44＝384(种)；③所取标号为3,3,2,2，共有排法N3＝A44＝24(种)．所以共有排法种数N＝N1＋N2＋N3＝384＋24×2＝432(种)．故选C.

答案： C

二、填空题

7．(2010·天津和平)有5名男生和3名女生，从中选出5人分别担任语文、数学、英语、物理、化学学科的科代表，若某女生必须担任语文科代表，则不同的选法共有________种(用数字作答)．

解析：由题意知，从剩余7人中选出4人担任4个学科科代表，共有A47＝840种．答案：840

8．在连续自然数100,101,102，…，999中，对于{0,1,2,3,4,5,6,7,8,9}，取三个不同且不相邻的数字按递增或递减的顺序排成的三位数有________个．

解析：分两类：①递减时，若有0，则0在个位，符合要求，从10个数字中选3个不相邻数字，相当于从10个位置中选3个不相邻的位置，故可将所选的3个位置插在其余7个位置的空位之中，故不同的情况共有C38种；②递增时，不能有0，则应从1到9的9个数字中，选3个不相邻的数字，同①有C37种，故所求的三位数有：C38＋C37＝91(个)．答案：91

9．某公司计划在北京、上海、兰州、银川四个候选城市投资3个不同的项目，且在同一个城市投资的项目不超过2个，则该公司不同的投资方案种数是________种．(用数字作答)

解析：由题意知按投资城市的个数分两类：①投资3个城市即A34种．②投资2个城市即C23A24种．共有不同的投资方案种数是A34＋C23A24＝60种．

答案：60

三、解答题

10．按下列要求分配6本不同的书，各有多少种不同的分配方式？

(1)分成三份，1份1本，1份2本，1份3本；

(2)甲、乙、丙三人中，一人得1本，一人得2本，一人得3本．

【解析方法代码108001142】

解析：(1)无序不均匀分组问题．先选1本有C16种选法；再从余下的5本中选2本有C25种选法；最后余下3本全选有C33种选法．故共有C16C25C33＝60种不同的分配方式．

(2)有序不均匀分组问题．由于甲、乙、丙是不同三人，在第(1)题的基础上，还应考虑再分配，故共有C16C25C33A33＝360种不同的分配方式．

11．(1)以AB为直径的半圆上，除A、B两点外，另有6个点，又因为AB上另有4个点，共12个点，以这12个点为顶点共能组成多少个四边形？

(2)在角A的一边上有五个点(不含A)，另一边上有四个点(不含A)，由这十个点(含A)可构成多少个三角形？

解析：(1)分类讨论：A、B只含有一个点时，共有2(C36＋C26C14)＝160个；

既含A又含B时，共有C26＝15个；

既不含A也不含B时，共有C410－1－C34C16＝185个．

所以共有160＋15＋185＝360个．

(2)含A点时，可构成C15C14＝20个三角形；

不含A点时，可构成C25C14＋C15C24＝70个三角形．

故共有20＋70＝90个三角形．

12．六人按下列要求站一排，分别有多少种不同的站法？

(1)甲不站两端；(2)甲、乙必须相邻；

(3)甲、乙不相邻；(4)甲、乙之间恰间隔两人．

【解析方法代码108001143】

解析：(1)方法一：要使甲不站在两端，可先让甲在中间4个

位置上任选1个，有A14种站法，然后其余5人在另外5个位置上作全排列，有A55种站

法，根据分步乘法计数原理，共有A14·A55＝480种站法．

方法二：若对甲没有限制条件共有A66种站法，甲在两端共有2A55种站法，从总数中减去这两种情况的排列数即得所求的站法数，共有A66－2A55＝480种站法．

(2)方法一：先把甲、乙作为一个“整体”，看作一个人，有A55种站法，再把甲、乙进行全排列，有A22种站法，根据分步乘法计数原理，共有A55·A22＝240种站法．方法二：先把甲、乙以外的4个人作全排列，有A44种站法，再在5个空档中选出一个供甲、乙站，有A15种站法，最后让甲、乙全排列，有A22种方法，共有A44·A15·A22＝240种站法．

(3)方法一：因为甲、乙不相邻，所以可用“插空法”．第一步，先让甲、乙以外的4个人站队，有A44种站法；第二步，再将甲、乙排在4人形成的5个空档(含两端)中，有A25种站法，故共有A44·A25＝480种站法．

方法二(间接法)：6个人全排列有A66种站法，由(2)知甲、乙相邻有A55·A22＝240种站法，所以不相邻的站法有A66－A55·A22＝720－240＝480(种)．

(4)方法一：先将甲、乙以外的4个人作全排列，有A44种站法，然后将甲、乙按条件插入站队，有3A22种站法，故共有A44·3A22＝144种站法．

方法二：先从甲、乙以外的4个人中任选2人排在甲、乙之间的两个位置上，有A24种；然后把甲、乙及中间2人看作一个“大”元素与余下2人作全排列，有A33种站法；最后对甲、乙进行全排列，有A22种站法，故共有A24·A33·A22＝144种站法．

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