节: 带有重复的子集和变异的变换 | 12.4 具有重复性的子集和变异的变异

章节大纲

Twelve violin students are competing for three seats in the orchestra: 1st Chair, 2nd Chair, and 3rd Chair. How many ways can a 1st Chair, 2nd Chair, and 3rd Chair be selected from the 12 competing students?
::十二位小提琴学生在乐团竞争三席:第一主席、第二主席、第三主席。从12位相互竞争的学生中选出第一主席、第二主席、第三主席可选用多少方法?

Permutations
::差异

Sometimes we want to order a select subset of a group. For example, suppose we go to an ice cream shop that offers 15 flavors. If we want to layer 3 scoops of different flavors on an ice cream cone, how many arrangements are possible? Here, the order matters so a chocolate, strawberry and vanilla cone is different than a strawberry, vanilla and chocolate cone. This is an example of permutations of a subset. We don’t need to know how many ways we can order all 15 flavors, just three choices. You have actually solved problems like this already using the . There are 15 choices for scoop one, 14 choices for scoop two and 13 choices for scoop 3, so $\begin{aligned} 15 \times 14 \times 13 = 2730 \end{aligned}$ .
::有时,我们想要订购一组的某个子集。例如, 假设我们去一家冰淇淋店, 提供15种口味。如果我们想在冰淇淋锥上堆放3个不同口味的勺子, 有多少安排是可能的? 这里, 顺序问题在于巧克力、草莓和香草香草锥子与草莓、香草和巧克力锥子不同。这是子酱、香草和巧克力锥子的变异例子。我们不需要知道我们有多少种方法可以订购所有15种口味, 只有3种选择。您实际上已经解决了类似问题了。第一次有15种选择, 第二次有14种, 第3次有13种选择, 所以 15x14x13=2730 。

We can use to solve this as well. Consider the expression: $\begin{aligned} \frac{15 \times 14 \times 13 \times 12!}{12!} = \frac{15!}{(15 - 3)!} = \frac{n!}{(n - r)!} \end{aligned}$ , where $\begin{aligned} n \end{aligned}$ represents the total number of elements in the set and $\begin{aligned} r \end{aligned}$ represents the number of elements in the subset we are selecting. Mathematically, this can be written using the notation $\begin{aligned} _{15} P_{3} \end{aligned}$ or $\begin{aligned} _{n} P_{r} \end{aligned}$ . To summarize, if we wish to find the number of permutations of $\begin{aligned} r \end{aligned}$ elements selected from a larger set containing $\begin{aligned} n \end{aligned}$ elements, we can use the formula: $\begin{aligned} _{n} P_{r} = \frac{n!}{(n - r)!} \end{aligned}$ .
::我们可以同时解决这个问题。考虑表达式 : 15x14x13x13x12! 12! =15! (15- 3) ! =n! (n-r) ! n 表示集元素的总数, r 表示我们所选择子集元素的数量。从数学角度讲, 这可以使用 15P3 或 nPr 来写入。概括地说, 如果我们想要找到从包含 n 元素的更大组中选择的 r 元素的变换数, 我们可以使用公式: nPr=n! (n-r) !!

In a set with $\begin{aligned} n \end{aligned}$ elements, in which $\begin{aligned} n_{1}, n_{2}, n_{3}, \dots \end{aligned}$ are indistinguishable, we can find the number of unique permutations of the $\begin{aligned} n \end{aligned}$ elements using the formula:
::在带有 n 元素的一组中, n1,n2,n3... 是无法区分的, 我们可以用公式找到 n 元素的独特变换数 :

$\begin{aligned} \frac{n!}{n_{1}! \times n_{2}! \times n_{3}! \times \dots} \end{aligned}$

::不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不

$\begin{aligned} ^{*} \end{aligned}$ Note that some textbooks use the notation $\begin{aligned} P_{r}^{n} \end{aligned}$ to represent $\begin{aligned} _{n} P_{r} \end{aligned}$ .
::* 注意一些教科书使用注解号表示nPr。

We can evaluate this expression easily on the calculator as well. First, enter the value of $\begin{aligned} n (15) \end{aligned}$ , then go to MATH $\begin{aligned} \to \end{aligned}$ PRB, select 2: $\begin{aligned} _{n} P_{r} \end{aligned}$ . Now enter the value of $\begin{aligned} r (3) \end{aligned}$ to get the expression 15 $\begin{aligned} _{n} P_{r} \end{aligned}$ 3 on your screen. Press ENTER one more time and the result is 2730.
::也可以在计算器上轻松评估这个表达式。首先, 输入 n( 15) 的值, 然后转到 MATHPRB, 选择 2: nPr。现在输入 r(3) 的值, 以便在屏幕上获取 15 nPr 3 的表达式。再按 ENTER 一次, 结果为 2730 。

Let's solve the following problems using permutations.
::让我们用变换方法解决以下的问题。

How many ways can a President, Vice President, Secretary and Treasurer be selecting from a club with ten members?
::一名主席、副主席、秘书和财务主任从有10名成员的俱乐部中挑选出多少种方式?

In the selection process here, the order matters so we are calculating the number of permutations of a subset of 4 members of the 10 member club. So,
::在此甄选过程中,顺序很重要, 所以我们在计算10个成员俱乐部的4名成员的分组数目。所以,

$\begin{aligned} _{10} P_{4} = \frac{10!}{(10 - 4)!} = \frac{10 \times 9 \times 8 \times 7 \times 6!}{6!} = 10 \times 9 \times 8 \times 7 = 5040. \end{aligned}$

::10P4=10!! (10- 4) =10x9x8x7x6! 6x10x9x8x7=5040 。

Consider the word, VIRGINIA. How many unique ways can these letters be arranged?
::考虑一下这个词,维吉尼亚,这些字母可以安排多少个独特的方式?

There are eight letters which can be arranged $\begin{aligned} 8! \end{aligned}$ ways. However, some of these arrangements will not be unique because there are multiple I’s in the word VIRGINIA. For example, if we let the three I’s be different colors, then we can see that there are several indistinguishable ways the I’s can be arranged.
::有8个字母可以排列 8 种方式。但是,有些这样的安排并不是独一无二的,因为Virginia 这个词里有多种我。比如,如果我们让三个我不同颜色,那么我们就能看到,有几种无法区分的方式可以安排我。

$\begin{aligned} V I R G I N I A, V I R G I N I A, V I R G I N I A, V I R G I N I A, V I R G I N I A, V I R G I N I A \end{aligned}$

::维吉尼亚、维吉尼亚、维吉尼亚、维吉尼亚、维吉尼亚、维吉尼亚、维吉尼亚、维吉尼亚

In fact, there are $\begin{aligned} 3! \end{aligned}$ or 6 ways that the three I’s can be arranged that are indistinguishable when the arrangement of the remaining letters is constant. To figure out the number of unique arrangements of the 8 letters with 3 that are indistinguishable we can find the permutations of the 8 letters and divide by the permutations of the indistinguishable items .
::事实上,有三种或六种方法可以安排我的三个字母,在其余字母的排列不变时,它们是无法区分的。要弄清楚8个字母和3个字母的独特安排有多少是无法区分的,我们能找到8个字母的变异和因无法区分的项目的变异而分裂的。

$\begin{aligned} \frac{8!}{3!} = \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3!}{3!} = 8 \times 7 \times 6 \times 5 \times 4 = 6720 \end{aligned}$

Consider the word PEPPERS. How many unique arrangements can be made of these letters?
::考虑一下PEPPERS这个字。这些字母可以做出多少个独特的安排?

There are seven letters in total, three of which are P and two of which are E. We can expand upon what we did in the last example and divide by the number of ways each of these letters can be arranged.
::共有7个字母,其中3个是P,2个是E。我们可以扩展我们在最后一个例子中所做的一切,并根据这些字母中每个字母的排列方式来划分。

$\begin{aligned} \frac{7!}{3! 2!} = \frac{7 \times 6 \times 5 \times 4 \times 3!}{3! \times 2 \times 1} = \frac{7 \times 6 \times 5 \times (2 \times 2)}{2} = 7 \times 6 \times 5 \times 2 = 420 \end{aligned}$

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the number of ways ways the 1st Chair, 2nd Chair and 3rd Chair can be selected from the 12 competing students.
::早些时候,有人要求你从12名相互竞争的学生中选出第一主席、第二主席和第三主席。

In the orchestra selection process, the order matters so we are calculating the number of permutations of a subset of 3 members of the 12 member club. So,
::在乐团的选拔过程中,顺序很重要所以我们在计算12个会员俱乐部的3名成员的分组数目。所以,

$\begin{aligned} _{12} P_{3} = \frac{12!}{(12 - 3)!} = \frac{12 \times 11 \times 10 \times 9!}{9!} = 12 \times 11 \times 10 = 1320 \end{aligned}$

::12P3=12! (12) =12x11x10x9! 9! =12x11x10=1320

Therefore, the orchestra slots can be filled in 1320 ways.
::因此,交响乐团的空位可以以1320的方式填补。

Example 2
::例2

Find $\begin{aligned} _{10} P_{6} \end{aligned}$ .
::查找 10P6。

$\begin{aligned} _{10} P_{6} = \frac{10!}{(10 - 6)!} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!} = 151, 200 \end{aligned}$

::10P6=10! (10- 6)!=10x9x8x7x6x5x4! 4!=151200

Example 3
::例3

On a team of 12 players, how many ways can the coach select players to receive on of each of the following awards (one award per player): most valuable player, best sportsmanship, most improved player.
::在一个由12名运动员组成的团队中,教练们能够从以下每一奖项(每个运动员一个奖项)中挑选出多少种方式:最有价值的运动员、最优秀的运动员、最优秀的运动员、最优秀的运动员。

$\begin{aligned} _{12} P_{3} = \frac{12!}{(12 - 3)!} = \frac{12 \times 11 \times 10 \times 9!}{9!} = 1, 320 \end{aligned}$

::12P3=12! (12) =12x11x10x9! 9! =1,320

Example 4
::例4

How many ways can 5 yellow, 4 red and 3 green balls be arranged in a row?
::5个黄色 4个红色 3个绿色球连续排列多少种方式?

$\begin{aligned} \frac{12!}{5! \times 4! \times 3!} & = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5! \times 4! \times 3!} \\ = \frac{12 \times 11 \times 10 \times 9 \times (4 \times 2) \times 7 \times 6}{(4 \times 3) \times 2 \times 3 \times 2} \\ = 11 \times 10 \times 9 \times 4 \times 7 \\ = 27, 720 \end{aligned}$

Review
::回顾

Evaluate the following expressions.
::评估以下表达式。

$\begin{aligned} _{8} P_{5} \end{aligned}$
::8P5 8P5

$\begin{aligned} _{11} P_{8} \end{aligned}$
::11P8 11P8

Evaluate and explain the results of each of the following: $\begin{aligned} _{5} P_{5},_{5} P_{0},_{5} P_{1} \end{aligned}$
::评估并解释以下各项的结果:5P5、5P0、5P1

Sarah needs to go to five different stores. How many ways can she go to two of them before lunch?
::Sarah需要去五家不同的商店她午餐前能去两家店吗?

In a race there are eleven competitors in a particular age group. How many possible arrangements are there for the top five finishers in this age group?
::在一个种族中,某一年龄组有11名竞争者,这一年龄组的前5名选手有多少可能的安排?

How many ways can eight distinct raffle prizes be awarded to fifteen ticket holders.
::向15名持票人颁发8个不同的奖项,

In a class of 24 students, there are six groups of four students. How many ways can a teacher select one group for each of three classroom maintenance responsibilities?
::在一个由24名学生组成的班级中,有六组四名学生。教师可以用多少方法为3个教室的维护责任各选择一个班级?

At a birthday party there are 6 unique prizes to be given randomly to the 6 quests such that no one guest receives more than one prize. How many ways can this be done?
::在生日派对上,有6个独特的奖项可以随机颁发给6个追求,这样,任何一位客人都得不到一个以上的奖项。有多少方法可以做到这一点?

How many unique ways can the letters in MISSISSIPPI be arranged?
::MISSIPIPI中的信件可以安排多少个独特的方式?

How many ways can two Geometry books, eight Algebra books and three Pre-Calculus books be arranged on a shelf if all the books of each respective subject are identical.
::两本几何学书、八本代数书和三本学前书,如果每个科目的所有书籍相同,可以用多少种方式放在一个架子上。

At a math department luncheon, the department chair has three $20 gift certificates to the local coffee shop, five $25 gift certificates to a local bookstore and two state of the art calculators to give as prizes amongst the 10 department members. If each teacher receives one prize, how many unique distributions of prizes can be made?
::在数学系午餐会上,系主任有三张20美元的当地咖啡店礼品券、五张25美元的当地书店礼品券和两张作为10名系成员奖品的艺术计算器。如果每名教师获得一个奖品,那么可以发放多少独特的奖品?

Review (Answers)
::回顾(答复)

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。

章节大纲

Permutations ::差异

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Permutations
::差异

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)