变差和组合

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In order to compute the probability of an event , you need to know the number of outcomes in the sample space and the number of outcomes in the event . Sometimes, determining the number of outcomes takes some work! Here, you will look at three techniques for counting outcomes.
::要计算事件概率, 您需要知道样本空间的结果数量和事件结果数量。 有时, 确定结果数量需要做一些工作 。 在此, 您将查看计算结果的三种技术 。

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Technique #1: The : Use this when there are multiple , each with their own outcomes, and you want to know how many outcomes there are for all the events together .
::技术##1: 技术: 当有多重时使用此功能, 每个都有自己的结果, 您想知道所有活动一起有多少结果 。

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At the local ice cream shop, there are 5 flavors of homemade ice cream -- vanilla, chocolate, strawberry, cookie dough, and coffee. You can choose to have your ice cream in a dish or in a cone . How many possible ice cream orders are there?
::在当地的冰淇淋店里,有五种自制的冰淇淋 香草、巧克力、草莓、饼干面团和咖啡。你可以选择把冰淇淋放在盘子或锥子上。有多少冰淇淋订单?

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If you list them all out, you will see that there are 10 ice cream orders . For each of the 5 flavors, there are 2 choices for how the ice cream is served (dish or cone). $\begin{align*}5 \cdot 2=10\end{align*}$
::如果你把它们都列出来, 你会看到有10个冰淇淋订单。 对于5种口味中的每一口味, 冰淇淋的配方( 甜甜或甜甜) 有2个选择。 52=10

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This idea generalizes to a principle called the Fundamental Counting Principle:
::这个概念概括了一项称为 " 基本计算原则 " 的原则:

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Fundamental Counting Principle: For independent events  $\begin{align*}A\end{align*}$ and $\begin{align*}B,\end{align*}$  if there are $\begin{align*}n\end{align*}$ outcomes in event  $\begin{align*}A\end{align*}$ and $\begin{align*}m\end{align*}$ outcomes in event $\begin{align*}B\end{align*}$ , then there are $\begin{align*}n \cdot m\end{align*}$ outcomes for events  $\begin{align*}A\end{align*}$ and  $\begin{align*}B\end{align*}$ together. 
::基本计算原则:对于独立事件A和B,如果事件A和事件B的结果为n,那么事件A和事件B的结果为n。

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The Fundamental Counting Principle works similarly for more than two events - multiply the number of outcomes in each event together to find the total number of outcomes.
::基本计数原则对两个以上事件同样适用 -- -- 将每个事件的结果数目相乘,以找出结果的总数。

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Technique #2: Permutations: Use this when you are counting the number of ways to choose and arrange a given number of objects from a set of objects .
::## 2 技术 2: 变换 : 当您在计算选择和排列一组对象中指定数量对象的方法数时使用此选项 。

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Your sister's 3rd grade class with 28 students recently had a science fair. The teacher chose 1st, 2nd, and 3rd place winners from the class. In how many ways could she have chosen the 1st, 2nd, and 3rd place winners?
::你姐姐的三年级班最近有28名学生参加了科学博览会。 老师从班上选择了第一、第二和第三位的优胜者。 她从多少方面选择了第一、第二和第三位的优胜者?

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This is called a problem because it is asking for the number of ways to choose and arrange 3 students from a set of 28 students.
::这被称为一个问题,因为它要求从一组28名学生中选择和安排3名学生的方法。

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In this problem, event  $\begin{align*}A\end{align*}$ is the teacher choosing 1st place, event  $\begin{align*}B\end{align*}$ is the teacher choosing 2nd place after 1st place has been chosen, and event  $\begin{align*}C\end{align*}$ is the teacher choosing 3rd place after 1st and 2nd place have been chosen.
::在此问题上,事件A是教师选择第1位,事件B是教师选择第1位之后的第2位,事件C是教师选择第1位之后的第3位。

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• Event  $\begin{align*}A\end{align*}$ has 28 outcomes because there are 28 students in the class. The teacher has 28 choices for 1st place.
  ::A活动有28个结果,因为班级有28名学生,教师有28个第1位选择。
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• Event  $\begin{align*}B\end{align*}$ has 27 outcomes because once 1st place has been chosen, there are 27 students left in the class that could get 2nd place.
  ::B事件有27个结果, 因为一旦第一位被选中, 班级中还有27个学生,
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• Event  $\begin{align*}C\end{align*}$ has 26 outcomes because once 1st place and 2nd place have been chosen, there are 26 students left in the class that could get 3rd place.
  ::C事件有26个结果, 因为一旦第1位和第2位被选中,

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By the Fundamental Counting Principle, the teacher has $\begin{align*}28 \cdot 27 \cdot 26=19656\end{align*}$  ways in which she could choose the 3 winners. Note that:
::根据 " 基本计数原则 " ,教师可以选择3名优胜者,其选择方式为282726=1956。

$\begin{align*}28 \cdot 27 \cdot 26=\frac{28 \cdot 27 \cdot 26 \cdot \bcancel{25 \cdot 24 \cdot 23 \cdots 3 \cdot 2 \cdot 1}}{\bcancel{25 \cdot 24 \cdot 23 \cdots 3 \cdot 2 \cdot 1}}=\frac{28!}{25!}=\frac{28!}{(28-3)!}\end{align*}$

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This is the idea behind the permutation formula. ( Recall that the symbol, !, means to multiply every whole number up to and including that whole number together. For example, $\begin{align*}5!=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1.\end{align*}$ )
::这就是变换公式背后的想法。 (回顾符号, !, 意指将每个整数乘在一起, 包括整个数。 例如, 5! =5+4+3+2+1 。 )

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Permutation Formula: The number of ways to choose and arrange $\begin{align*}k\end{align*}$  objects from a group of $\begin{align*}n\end{align*}$  objects is:
::变异公式:选择和排列 n 组对象的 k 对象的方法数为:

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$\begin{align*}_nP_k=\frac{n!}{(n-k)!}\end{align*}$
::NPKKN!

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Technique #3: Combinations: Use this when you are counting the number of ways to choose a certain number of objects from a set of objects (the order/arrangement of the objects doesn't matter) . 
::技术 # 3 : 组合 : 当您在计算从一组对象中选择一定数量对象的方法数时使用此选项( 对象的顺序/ 排列并不重要 ) 。

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A teacher has a classroom of 28 students, she wants 3 of them to do a presentation, and she wants to know how many choices she has for the three students. 
::一名教师有28名学生的教室, 她想让其中3人做演讲, 她想知道她为这3名学生做了多少选择。

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This is called a problem because it is asking for the number of ways to choose 3 students from a set of 28 students. With combinations, the order doesn't matter. Choosing Bobby, Sarah and Matt for the presentation is the same as choosing Sarah, Bobby and Matt for the presentation.
::这是个问题,因为它要求从一组28名学生中选择3名学生的方法。如果组合,命令并不重要。选择Bobby、Sarah和Matt参加演讲与选择Sarah、Bobby和Matt参加演讲相同。

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From permutations, you know that there are 19656 ways to choose and arrange 3 students from the class of 28. This calculation will be counting each group of 3 people more than once. How many times is each group of three being counted? From permutations, 3 chosen people can be arranged in  $\begin{align*}_3P_3\end{align*}$ ways.
::从变换中,你知道有19656年6月的方法从28年级中选择和安排3名学生。这个计算方法将计算每组3人不止1次。每组3人要计算多少次?从变换中,3人可以按3P3方式安排。

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$\begin{align*}_3P_3=\frac{3!}{(3-3)!}=\frac{3!}{0!}=\frac{3!}{1}=3 \cdot 2 \cdot 1=6 \ \text{ways}.\end{align*}$
::3P3=3! (3- 3)!=3. 0!=3! 1=3Q2=1=6方法。

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This means that every group of three has been counted 6 times in the 19656 calculation. To determine the number of ways the teacher could choose 3 students where the order doesn't matter, take 19656 and divide by 6.
::这意味着在19656年的计算中,每组三人就计算了6次。为了确定教师选择3名学生的方式数量,如果顺序无关紧要,则采用19656,除以6。

$\begin{align*}\frac{19656}{6}=3276\end{align*}$   $\begin{align*}\frac{19656}{6}=3276\end{align*}$

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The teacher has 3276 choices for the three students to make a presentation. 
::教师有3276个选择,供3名学生作演讲。

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In general, the permutation formula can be turned into the combination formula by dividing by the number of ways to arrange $\begin{align*}k\end{align*}$  objects, which is $\begin{align*}k!.\end{align*}$
::一般而言,变异公式可以通过除以 k 对象的排列方式,即 k! 来转换成组合公式。

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Combination Formula: The number of ways to choose $\begin{align*}k\end{align*}$  objects from a group of $\begin{align*}n\end{align*}$  objects is:
::组合公式:从一组 n 对象中选择 k 对象的方法数为 :

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$\begin{align*}_nC_k=\frac{_nP_k}{k!}=\frac{n!}{k!(n-k)!}\end{align*}$
::NCK=nPKK! NKK! (nKK) ! (nKK) ! (nKKK) ! (nKKK) !

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In the problems below you will see how to use the fundamental counting principle, permutations, and combinations to help you compute probabilities. Note that whenever you can use permutations you can also use the fundamental counting principle, because the permutation formula is derived from the fundamental counting principle.
::在下面的问题中,您将看到如何使用基本计算原则、变换和组合来帮助您计算概率。请注意,只要您可以使用变位,您也可以使用基本计算原则,因为变位公式来自基本计算原则。

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Using the Fundamental Counting Principle
::使用基本计算原则

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Suppose you are ordering a sandwich at the deli. There are 5 choices for bread, 4 choices for meat, 12 choices for vegetables, and 3 choices for a sauce. How many different sandwiches can be ordered? If you choose a sandwich at random, what's the probability that you get turkey and mayonnaise on your sandwich?
::假设您在熟食店点了三明治。 面包有5种选择,肉有4种选择,蔬菜有12种选择,酱有3种选择。 可以订购多少种不同的三明治? 如果你随机选择三明治,那么在三明治上吃火鸡和蛋黄酱的概率是多少?

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In order to answer this probability question you need to know:
::为了回答这个概率问题 你需要知道:

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1. The total number of sandwiches that can be ordered.
   ::可以订购的三明治总数。
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2. The number of sandwiches that can be ordered that involve turkey and mayonnaise.
   ::可以订购的涉及火鸡和蛋黄酱的三明治数量。

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In each case, you can use the fundamental counting principle to help.
::在每种情况下,您都可以使用基本计算原则提供帮助。

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1. A sandwich is made by choosing a bread, a meat, a vegetable, and a sauce. There are 5 outcomes for the event of choosing bread, 4 outcomes for the event of choosing meat, 12 outcomes for the event of choosing vegetables, and 3 outcomes for the event of choosing a sauce. The total number of sandwiches that can be ordered is:  $\begin{align*}& = (\text{Choices of Bread}) \cdot (\text{Choices of Meat}) \cdot (\text{Choices of Vegetables}) \cdot (\text{Choices of Sauce}) \\ & = 5 \cdot 4 \cdot 12 \cdot 3 \\ & = 720 \\\end{align*}$  
   ::三明治是通过选择面包、肉、蔬菜和酱汁来做的。 选择面包有5个结果,选择肉有4个结果,选择蔬菜有12个结果,选择酱有3个结果。 可以订购的三明治总数是:==(面包的巧克力)、(肉的甜菜)、(蔬菜的甜菜)=54123=720。
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2. A sandwich with turkey and mayonnaise is made by choosing a bread, turkey, a vegetable, and mayonnaise. There are 5 outcomes for the event of choosing bread, there is 1 outcome for the event of choosing turkey, there are 12 outcomes for the event of choosing vegetables, and there is 1 outcome for the event of choosing mayonnaise. The total number of sandwiches with turkey and mayonnaise that can be ordered is: $\begin{align*}& = (\text{Choices of Bread}) \cdot (\text{Meat : Turkey}) \cdot (\text{Choices of Vegetables}) \cdot (\text{Sauce : Mayonnaise}) \\ & = 5 \cdot 1 \cdot 12 \cdot 1 \\ & = 60 \\\end{align*}$  
   ::配有火鸡和蛋黄酱的三明治是通过选择面包、火鸡、蔬菜和蛋黄酱做的。选择面包有5个结果,选择火鸡有1个结果,选择蔬菜有12个结果,选择蛋黄酱有1个结果。可以订购的配有火鸡和蛋黄酱的三明治总数是:=(面包的巧克力)(Meat:土耳其)(蔬菜的巧克力)(Sauce:Mayonnaise)=51121=60。

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The probability of a sandwich with turkey and mayonnaise is:
::配有火鸡和蛋黄酱的三明治的概率是:

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$\begin{align*}\frac{\text{Number of sandwiches that can be ordered with turkey and mayonnaise}}{\text{Total number of sandwiches that can be ordered}} = \frac{60}{720} = \frac{1}{12}\end{align*}$
::可以用火鸡和蛋黄酱订购的三明治数量

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Using Permutations 
::使用变换

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In your class of 35 students, there are 20 girls and 15 boys. There is a class competition and 1st, 2nd and 3rd place winners are decided. What is the probability that all of the winners are boys?
::在你的35个学生班里,有20个女孩和15个男孩。有一个班级竞赛,第1、第2和第3位获胜者被决定。所有获胜者都是男孩的可能性有多大?

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In order to answer this probability question you need to know:
::为了回答这个概率问题 你需要知道:

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1. The total number of 1st, 2nd, 3rd place winners that can be chosen.
   ::排名第1、第2、第3位的获奖者总数
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2. The number of 1st, 2nd, 3rd place winners that are all boys that can be chosen.
   ::排名第一、第二、第三的获胜者 全部都是可以被选中的男孩

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In each case, you are dealing with permutations, because the order of the people for 1st, 2nd and 3rd place matters .
::在每种情况下,你处理的是变异, 因为第1、第2和第3位的人的顺序很重要。

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1. There are 35 students and 3 need to be chosen and arranged into 1st, 2nd and 3rd place.
::1. 共有35名学生,3名学生需要挑选,并排在第1、第2和第3位。

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The number of ways 3 winners can be chosen: 
::选择3位获奖者的方式数量 :

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$\begin{align*}& = \ _{35}P_3 = \frac{35!}{(35-3)!}\\ & = \frac{35!}{32!} \\ & = \frac{35 \cdot 34 \cdot 33 \cdot \bcancel{32 \cdot 31 \cdot \cdot \cdot 1}}{\bcancel{32 \cdot 31 \cdot 30 \cdot 29 \cdot 28 \cdot \cdot \cdot 1}} \\ & = 35 \cdot 34 \cdot 33 \\ & = 39270\end{align*}$
::35P3=3535! (35- 3)!=35. 32!=35. 32!

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2. There are 15 boys and 3 need to be chosen and arranged.
::2. 有15名男孩和3名男孩需要挑选和安排。

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The number of ways all boy winners can be chosen: 
::选择所有男孩赢家的方式数量 :

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$\begin{align*}& = \ _{15}{P}_3 = \frac{15!}{(15-3)!} \\ & = \frac{15 !}{12!} \\ & = \frac{15 \cdot 14 \cdot 13 \cdot \bcancel{12 \cdot 11 \cdot \cdot \cdot 3 \cdot 2 \cdot 1}}{\bcancel{12 \cdot 11 \cdot \cdot \cdot 3 \cdot 2 \cdot 1}} \\ & = 15 \cdot 14 \cdot 13 \\ & = 2730\end{align*}$  
::15P3=1515! (15- 3)!=15!12!=15. 12!

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The probability of all boy winners is: 
::所有男生获胜的概率是:

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$\begin{align*}\frac{\text{Number of ways all boys can be chosen}}{\text{Number of ways 3 winners can be chosen}} = \frac{2730}{39270} \approx 7 \%\end{align*}$  
::3名获奖者被选为273039270=7%

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Now, let's look at some problems involving combinations. 
::现在,让我们来看看一些 涉及组合的问题。

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In your class of 35 students, there are 20 girls and 15 boys. 5 students are chosen at random for a presentation. What is the probability that the group is made of all boys?
::在你的35名学生班中,有20名女生和15名男生。 5名学生是随机挑选的,参加演讲。 这个群体由所有男生组成的可能性有多大?

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In order to answer this probability question you need to know:
::为了回答这个概率问题 你需要知道:

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1. The total number of groups that can be formed.
   ::可以组建的团体总数。
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2. The number of groups with all boys.
   ::所有男孩的组数。

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In each case, you are dealing with combinations, because the order of the people for the presentation doesn't matter .
::在每种情况下,你处理的都是组合, 因为演讲的人的顺序无关紧要。

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1. There are 35 students in the class and 5 to be chosen. The number of ways the 5 could be chosen are:
::1. 班级有35名学生,5名待选,可选方式如下:

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$\begin{align*}_{35}C_5 = \frac{35!}{5!(35-5)!}=\frac{35!}{5!30!} &= \frac{35 \cdot 34 \cdot 33 \cdot 32 \cdot 31 \cdot \bcancel{30 \cdots 3 \cdot 2 \cdot 1}}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot \bcancel{30 \cdot 29 \cdot 28 \cdots 3 \cdot 2 \cdot 1}}\\ &= \frac{35 \cdot 34 \cdot 33 \cdot 32 \cdot 31}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\\ &= \frac{38955840}{120}\\ &= 324632\end{align*}$
::35C5=35355!5(35-5)!=355!5!30! 35C5=35C5=35C5=355!5(5-5)! 35C5=35C5=35C5=35C5=35C5=35C5=35C5=35C5=35C5=355!5!(35-5)! =35.5!5!30! 35C3=35C3}335C3=3333*32}331}30}3}3QA3}3}3Q1}3Q1}3Q1}315=3255840120=324632

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2. There are 15 boys in the class and 5 boys to be chosen. The number of ways the 5 could be chosen are:
::2. 班级中有15名男孩,5名男孩有待选择,选择5名男孩的方法如下:

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$\begin{align*}_{15}C_5 =\frac{15!}{5!(15-5)!}=\frac{15!}{5!10!} &= \frac{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\\ &= \frac{360360}{120}\\ &= 3003 \end{align*}$
::15C5=151515!5(15-5)!=15.5!5!10! 15C14113112-115}1154}3C1=360360120=3003

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The probability of an all boy group is:
::所有男孩群的概率如下:

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$\begin{align*}\frac{\text{Number of ways all boys group can be chosen}}{\text{Number of ways 5 students can be chosen}} = \frac{3003}{324632} \approx 0.9 \%\end{align*}$
::5名学生可被选为=3003324632+0.9%

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You are driving your friends to the beach in your car. Your car has room for 4 additional passengers besides yourself. You have 10 friends (not including yourself) going to the beach. 
::你开车载着你的朋友去海滩。你的汽车除了你还有4位乘客。你有10位朋友(不包括你自己)去海滩。

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The number of ways in which the friends who will ride in your car could be chosen is . 
::选择坐在你车里的朋友的方式是多少。

---

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Examples
::实例实例实例实例

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Example 1
::例1

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Suppose you draw two cards from a standard deck (one after the other without replacement) and record the results. 
::假设您从标准甲板上抽出两张牌(一张接一张不替换)并记录结果。

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1. How many outcomes are there?
   ::有多少成果?
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2. How many ways are there to choose an ace and then a four?
   ::有多少方法可以选择一个A,然后选择一个4?
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3. What is the probability that you choose an ace and then a four?
   ::你选择一个A然后选择一个4的概率是多少?

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a) This is an example of a permutation, because the order matters. You are choosing 2 cards from a set of 52 cards.
:a) 这是一个变换的例子,因为顺序很重要。您是从一组52张卡中选择两张卡。

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$\begin{align*}_{52}P_2=\frac{52!}{(52-2)!}=\frac{52!}{50!}=50 \cdot 51=2652\end{align*}$
::52P2=52! (52-2)!=52. 50!=5051=2652

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b) Choosing an ace and choosing a four are independent events. There are 4 aces and 4 fours. By the fundamental counting principle, there are $\begin{align*}4 \cdot 4=16\end{align*}$  ways to choose an ace and then a four.
:b) 选择一个A,选择一个A,选择一个4是独立事件,有4个A,4个4个4,根据基本计算原则,选择一个A,然后选择4个A,有4个4个16种方法。

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c) The probability that you choose an ace and then a four is $\begin{align*}\frac{16}{2652} \approx 0.6\%\end{align*}$ .
:c) 选择A后选择4的概率为1626520.6%。

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Calculate $\begin{align*}_{10}P_4\end{align*}$  and $\begin{align*}_{10}C_4.\end{align*}$  Interpret each calculation.
::计算 10P4 和 10C4 。 解释每次计算 。

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$\begin{align*}_{10}P_4=\frac{10!}{(10-4)!}=\frac{10!}{6!}=10 \cdot 9 \cdot 8 \cdot 7=5040.\end{align*}$  This is the number of ways to choose and arrange four objects from a set of 10 objects.
::10P4=10!! (10- 4)!=10. 6!=10987=5040。这是从一组 10 个对象中选择和排列四个对象的方法数 。

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$\begin{align*}_{10}C_4=\frac{10!}{4!(10-4)!}=\frac{10!}{4!6!}=\frac{10 \cdot 9 \cdot 8 \cdot 7}{4 \cdot 3 \cdot 2 \cdot 1}=\frac{5040}{24}=210.\end{align*}$  This is the number of ways to choose four objects from a set of 10 objects. 
::10C4=100. 4! (10- 4)!=10. 4! 6! =10Q8874321=504024=210。 这是从一组 10 个对象中选择四个对象的方法数 。

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Example 3
::例3

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Make up a probability question that could be solved with the calculation $\begin{align*}\frac{_3C_2}{_{15}C_2}.\end{align*}$
::得出一个概率问题,可以通过计算 3C215C2解决。

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The total number of outcomes in the sample space is $\begin{align*}_{15}C_2,\end{align*}$  which is the number of ways to choose 2 objects from a set of 15. The number of outcomes in the event that you are calculating the probability of is $\begin{align*}_3C_2,\end{align*}$  which is the number of ways to choose 2 objects from a set of 3. Here is one possible question:
::样本空间的结果总数为 15C2, 即从一组15中选择 2 个天体的方法数量。 如果计算出3C2 的概率,结果数量为 3C2, 从一组3中选择 2个天体的方法数量。

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The math club has 15 members. 12 are upperclassmen and 3 are freshman. 2 members of the club need to be chosen to make a morning announcement. What is the probability that the 2 who are chosen are both freshmen?
::数学俱乐部有15名成员。 12名是高年级学生,3名是大一。 需要挑选2名俱乐部成员来做晨报。 被选中的2名是新生的可能性有多大?

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CK-12 PLIX Interactive
::CK-12 PLIX 交互式互动

Summary

播放段落 A permutation is the number of ways to choose and arrange $\begin{align*}k\end{align*}$  objects from a group of n objects. (order matters)     $\begin{align*}_nP_k=\frac{n!}{(n-k)!}\end{align*}$ ::变换是指从一组 n 对象中选择和排列 k 对象的方法数。 (顺序事项) nPk=n! (n-k)! 播放段落 A combination is the number of ways to choose $\begin{align*}k\end{align*}$  objects from a group of n objects. (order doesn't matter)     $\begin{align*}_nC_k=\frac{n!}{k!(n-k)!}\end{align*}$   ::组合是从一组 n 对象中选择 k 对象的方法数 。 (顺序不重要) nCk=n!k! (n-k) ! 播放段落 The Fundamental Counting Principle states that if an event can be chosen in $\begin{align*}p\end{align*}$  different ways and another independent event can be chosen in $\begin{align*}q\end{align*}$  different ways, the number of different arrangements of the events is $\begin{align*}p\times q\end{align*}$   ::基本计数原则指出,如果以不同方式选择事件,而以q方式选择另一个独立事件,则不同事件安排的数目是pxq。

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Calculate each and interpret each calculation in words.
::每个计算单位并用单词解释每个计算单位。

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1. $\begin{align*}_8P_2\end{align*}$
::1. 8P2

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2. $\begin{align*}_8P_8\end{align*}$
::2. 8P8

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3. $\begin{align*}_8C_8\end{align*}$
::3. 8C8

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4. $\begin{align*}_{14}C_8\end{align*}$
::4. 14C8 14C8

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5. Will $\begin{align*}_nP_k\end{align*}$  always be larger than $\begin{align*}_nC_k\end{align*}$  for a given $\begin{align*}n,k\end{align*}$  pair?
::5. 对于给定n,k对,nPk是否总是大于nCk?

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6. Your graphing calculator has the combination and permutation formulas built in. Push the MATH button and scroll to the right to the PRB list. You should see  $\begin{align*}nPr\end{align*}$ and  $\begin{align*}nCr\end{align*}$ as options. In order to use these: 1) On your home screen type the value for $\begin{align*}n\end{align*}$ ; 2) Select  $\begin{align*}nPr\end{align*}$ or $\begin{align*}nCr\end{align*}$ ; 3) Type the value for  $\begin{align*}k\end{align*}$ ( $\begin{align*}r\end{align*}$ on the calculator). Use your calculator to verify that $\begin{align*}_{10}C_5=252\end{align*}$ .
::6. 您的图形计算计算器有组合和排列公式。 按下 MATH 按钮并滚动到 PRB 列表右侧。 您应该将 nPr 和 nCr 作为选项。 为了使用这些选项1) 在您的主屏幕上输入 n 的值;(2) 选择 nPr 或 nCr;(3) 输入 k (计算器上的 r) 值。使用您的计算器来验证 10C5=252 。

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First, state whether each problem is a permutation or combination problem. Then, solve.
::首先,指出每个问题是一个变异问题还是综合问题。然后,解决。

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7. Suppose you need to choose a new combination locker. You have to choose 3 numbers, each different and between 0 and 40. How many choices do you have for the combination? If you choose at random, what is the probability that you choose 0, 1, 2 for your combination?
::7. 假设您需要选择一个新的组合储物柜。 您必须选择 3 个数字, 每个数字不同, 介于 0 到 40 之间, 您对组合有多少选择? 如果您随机选择, 您选择 0, 1, 2 的组合概率是多少 ?

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8. You just won a contest where you can choose 2 friends to go with you to a concert. You have five friends (Amy, Bobby, Jen, Whitney, and David) who are available and want to go. If you choose two friends at random, what is the probability that you choose Bobby and David?
::8. 你刚刚赢得了一场比赛,在那里你可以选择两个朋友与你一起参加音乐会。你有五个朋友(艾米、鲍比、延、惠特尼和大卫)可以参加并愿意参加。如果你随机选择两个朋友,你选择鲍比和大卫的概率是多少?

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9. There are 12 workshops at a conference and Michael has to choose 4 to attend. In how many ways can he choose the 4 to attend?
::9. 会议有12个讲习班,迈克尔必须选择4个参加,他可以选择多少方式选择4个参加?

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10. 10 girls and 4 boys are finalists in a contest where 1st, 2nd, and 3rd place winners will be chosen. What is the probability that all winners are boys? 
::10. 10名女孩和4名男孩是第1、第2和第3位获胜者的决赛选手,所有获胜者都是男孩的可能性有多大?

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11. Using the information from the previous problem, what is the probability that all winners are girls?
::11. 利用上一个问题的信息,所有获胜者都是女孩的概率有多大?

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12. You visit 12 colleges and want to apply to 4 of them. 5 of the colleges are within 100 miles of your house. If you choose the colleges to apply to at random, what is the probability that all 4 colleges that you apply to are within 100 miles of your house? 
::你访问过12所学院,并想申请其中4所学院。5所学院离你家100英里以内。如果你选择了随机申请的学院,那么你申请的所有4所学院在离你家100英里以内的可能性是多少?

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13. For the 12 colleges you visited, you rank your top five. In how many ways could you do this? Your friend Jesse randomly tries to guess the five colleges that you choose and the order that you ranked them in. What is the probability that he guesses correctly?
::13. 对于你访问的12所学院,你将排名前5位。你能用多少种方式做到这一点?你的朋友杰西随机地试图猜测你选择的5所学院以及你排列这些学院的顺序。他猜对的概率是多少?

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14. For the special at a restaurant you can choose 3 different items from the 10 item menu. How many different combinations of meals could you get? If the waiter chooses your 3 items at random, what's the probability that you get the soup, the salad, and the pasta dish?
::14. 对于餐厅的特餐,您可以从10项菜单中选择3个不同的项目。您可以得到多少不同的膳食组合?如果服务员随机选择您的3项,那么您能得到汤、沙拉和意大利面条的概率是多少?

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15. In a typical poker game, each player is dealt 5 cards. A royal flush is when the player has the 10, Jack, Queen, King, and Ace all of the same suit . What is the probability of a royal flush? Hint: How many 5 card hands are there? How many royal flushes are there?
::15. 在典型的扑克游戏中,每个玩家都有5张牌。王牌是当玩家拥有10张,杰克、王后、国王和王牌,所有相同的西装。皇牌的概率是多少?提示:有多少5张牌?有多少皇牌?有多少皇牌?

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16. The board of directors of a firm has 10 members, They sit around a table that is in the shape of a U. How  many ways can they sit at the table? If they need to select a chairperson, a secretary, and a vice chair, how many ways can they select these members? If they had to add a treasurer, by how much will the offices held decrease?
::16. 公司董事会有10名成员,他们坐在一个以美国为形状的桌子上。他们能坐在桌子上坐多少种方式?如果他们需要挑选一名主席、一名秘书和一名副主席,他们可以选择多少种方式?如果他们必须增加一名司库,这些办公室将减少多少?

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17. Your school is trying to schedule 8 courses where each course can be offered only once. How many ways can the 8 courses be scheduled?
::17. 你们的学校正试图安排8个课程,每个课程只能提供一次,这8个课程可以安排多少种方式?

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18. You and your friend have reserved seats at the school musical. As well, two of your friends are also going and each taking a friend. How many ways can you sit if there were no seating restrictions? What if each of the couples wanted to sit together?
::18. 你和你的朋友在学校音乐会有预留座位,另外,两个朋友也要去,每个朋友都带朋友,如果没有座位限制,你还能坐几条路?如果每一对夫妇都想坐在一起呢?

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19. Eight people are seated at a round table where one person is seated by the window. How may possible arrangements of the people relative to the window are there?
::19. 8人坐在一个圆桌会议上,一人坐在窗口旁,人民与窗口相比可能做出何种安排?

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Review (Answers)
::审查(答复)

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Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。