Course: 8.4 与无法区分的成员的变动 | The Way To Learn

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与无法区分成员的差异
How many unique passwords can be created using the letters in the word “password”?
::使用“密码”一词中的字母可以创建多少个独特的密码?

Since the two “s” letters appear exactly the same, we cannot simply calculate the number of permutations of 8 letters. We also cannot simply calculate as if there were only seven letters, since there are eight locations in which to place letters.
::由于两个“s”字母看起来完全相同,我们无法简单地计算8个字母的变换次数,我们也不能简单地计算,似乎只有7个字母,因为有8个地方可以寄信。

By the time we return to this question after the lesson, you will know how to handle these types of permutations, and the answer should be clear.
::等我们吸取教训之后再讨论这个问题时,你就会知道如何处理这些类型的变相,答案应该很清楚。

Permutations with Indistinguishable Members
::与无法区分成员的差异

When a collection of units includes some members that appear the same, the number of permutations is reduced by the number of arrangements that result only from exchanging the identical members.
::当一个单位的集合包括一些看起来相同的成员时,变换次数将减少,只因交换相同成员而作的安排数目。

In order to exclude the number of permutations that are effectively the same due to identical members, we need to divide the number of possible permutations of all the items by the product of the of the number of indistinguishable members. As a formula, this looks like:
::为了排除因成员相同而实际上相同的变异数目,我们需要将所有项目可能的变异数目除以不可区分成员数目的产物。

$\begin{aligned} \frac{n!}{n_{1}! \times n_{2}! \dots \times n_{k}!} \end{aligned}$
$\begin{align*}\frac{n!}{n_1! \times n_2! \ldots \times n_k!}\end{align*}$
::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!

The explanation of why the division works is, unfortunately, beyond the scope of this book. At this point, I recommend learning the formula and applying it as needed, that is, whenever you need to calculate the permutations possible in a collection of items including some that are indistinguishable.
::不幸的是,解释为什么分部的工作超出了这本书的范围。此时此刻,我建议学习公式,并根据需要适用公式,也就是说,只要你需要计算一系列项目中可能的差异,包括一些无法区分的项目。

Rearranging Letters
::重新排列信函

How many ways can the letters ABBCCC be arranged so that the permutations are distinguishable?
::ABBCCCC能安排多少字母, 才能区分不同的字母?

Use the formula for calculating permutations with indistinguishable members: $\begin{align*}\frac{n!}{n_1! \times n_2! \ldots \times n_k!}\end{align*}$
::使用公式计算无法区分的成员的变异 : n! n1! xn2!... xnk!

There are six letters, so $\begin{align*}n=6\end{align*}$
::有六个字母, 所以 n=6

There is one “A”, so $\begin{align*}n_1=1\end{align*}$
::有一个“A”, 所以 n1=1

There are two “B’s”, so $\begin{align*}n_2=2\end{align*}$
::有两个 " B " , 所以 n2=2

There are three “C’s”, so $\begin{align*}n_3=3\end{align*}$
::有三个 " C " , 所以 n3=3

$\begin{aligned} \frac{6!}{1! \times 2! \times 3!} \\ \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(1) \times (2 \times 1) \times (3 \times 2 \times 1)} \\ \frac{720}{(1) \times (2) \times (6)} \\ \frac{720}{12} \\ 60 \end{aligned}$
$\begin{align*}& \qquad \quad \frac{6!}{1! \times 2! \times 3!}\\ & \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(1) \times (2 \times 1) \times (3 \times 2 \times 1)}\\ & \qquad \ \frac{720}{(1) \times (2) \times (6)}\\ & \qquad \qquad \ \frac{720}{12}\\ & \qquad \qquad \ \ 60\end{align*}$

There are 60 ways to arrange the letters ABBCCC that appear unique.
::ABBCCCC有60种方式来安排看起来独一无二的信

Arranging Shirt Colors
::排列衬衫颜色

A classroom has eight students in it. Three students are wearing red shirts, two are wearing white shirts, two are wearing blue shirts, and one is wearing a green shirt. If the students line up at the door for lunch break, how many ways could the shirt colors be arranged?
::一个教室里有8个学生。 3个学生穿着红色衬衫,2个身着白色衬衫,2个身着蓝色衬衫,1个身着绿色衬衫。如果学生们排在门前休息午餐,那么可以安排多少种方式的衬衫颜色?

There are eight students, so $\begin{align*}n = 8\end{align*}$
::有八个学生,所以N=8

There are three red shirts, so $\begin{align*}n_1 = 3\end{align*}$
::有三件红衬衫所以N1=3

There are two white shirts, so $\begin{align*}n_2 = 2\end{align*}$
::有两件白衬衫所以N2=2

There are two blue shirts, so $\begin{align*}n_3 = 2\end{align*}$
::有两件蓝色衬衫所以N3=2

The one green shirt is not a multiple, so we don’t need to worry about it, but it does not hurt to set $\begin{align*}n_4 = 1\end{align*}$ , since multiplying by one doesn’t change anything anyway.
::绿色衬衫不是一个多重, 所以我们不需要担心, 但设置 n4=1并不痛, 因为乘以乘以一也改变不了任何东西。

$\begin{aligned} \frac{8!}{(3!) (2!) (2!) (1!)} \\ \frac{40, 320}{(6) (2) (2) (1)} \\ \frac{40, 320}{24} \\ 3, 360 \end{aligned}$
$\begin{align*}& \frac{8!}{(3!)(2!)(2!)(1!)}\\ & \ \ \frac{40, 320}{(6)(2)(2)(1)}\\ & \quad \ \ \frac{40,320}{24}\\ & \qquad 3,360\end{align*}$

Creating Teams
::创建团队

Suppose there are twenty-five students in the auditorium and you are responsible for forming them into teams A, B, and C, with six students each, and team D of seven students. How many ways could the students be arranged?
::假设礼堂里有25名学生,你负责把他们组成A、B和C组,每个组有6名学生,D组有7名学生。可以安排多少方法?

This problem looks different, but the concept is the same, and you can use the same formula for calculation.
::这个问题看起来不同, 但概念是一样的, 您可以使用相同的公式来计算。

There are twenty-five students to choose from: $\begin{align*}n = 25\end{align*}$
::有25名学生可以选择:n=25

There are six students in team A: $\begin{align*}n_1 = 6\end{align*}$
::A组有6名学生: n1=6

There are six students in team B: $\begin{align*}n_2 = 6\end{align*}$
::B组有6名学生:N2=6

There are six students in team C: $\begin{align*}n_3 = 6\end{align*}$
::C组有6名学生: n3=6

There are seven students in team D: $\begin{align*}n_4 = 7\end{align*}$
::D组有7名学生:N4=7

$\begin{aligned} \frac{25!}{(6!) (6!) (6!) (7!)} \\ \frac{1.55 \times 10^{25}}{(720) (720) (720) (5040)} \end{aligned}$
$\begin{align*}& \quad \ \ \frac{25!}{(6!)(6!)(6!)(7!)}\\ & \frac{1.55 \times 10^{25}}{(720)(720)(720)(5040)}\end{align*}$

8, 245, 512, 475, 200 - Wow, that is a lot of possible teams!
::8,245,512,475,200 - 哇,这是很多可能的团队!

Earlier Problem Revisited
::重审先前的问题

How many unique passwords can be created using the letters in the word “password”?
::使用“密码”一词中的字母可以创建多少个独特的密码?

This one should be easy now, we’ll use the formula $\begin{align*}\frac{n!}{n_1! \times n_2! \ldots \times n_k!}\end{align*}$ :
::使用公式n!n1!

There are eight letters total, so $\begin{align*}n = 8\end{align*}$
::一共八个字母所以N=8

The only doubled letter is “s”, and it appears twice, so $\begin{align*}n_1 = 2\end{align*}$
::唯一翻一番的字母是“s”, 出现两次, 所以 n1=2

We don’t need to worry about the other letters, since they would each be 1
::我们不必为其他信件担心,

$\begin{aligned} \frac{8!}{2!} \\ \frac{40, 320}{2} \\ 20, 160 \end{aligned}$
$\begin{align*}& \quad \ \frac{8!}{2!}\\ & \frac{40,320}{2}\\ & \ 20,160\end{align*}$

Examples
::实例

Example 1
::例1

How many ways can the letters in the word "banana" be arranged?
::"香蕉"这个词里的信能安排多少种方式?

There are six letters, one has three multiples, one has two multiples, and one is singular:
::有六个字母,一个有三个倍数, 一个有两个倍数, 一个是单数 :

$\begin{aligned} \frac{6!}{(3!) (2!)} \\ \frac{720}{12} & = 60 a r r a n g e m e n t s \end{aligned}$
$\begin{align*}& \frac{6!}{(3!)(2!)}\\ \frac{720}{12} & =60 \ arrangements\end{align*}$
::6! 6! (3! ) (2!) 72012=60安排

Example 2
::例2

How many passwords can be made from the letters in "summers"?
::从“总和”中的字母可以密码多少?

There are seven letters, two of which each repeat twice:
::共有七封信,其中两封信各重复两次:

$\begin{aligned} \frac{7!}{(2!) (2!)} \\ \frac{5040}{4} & = 1260 p a s s w o r d s \end{aligned}$
$\begin{align*}& \frac{7!}{(2!)(2!)}\\ \frac{5040}{4}& =1260 \ passwords\end{align*}$
::7! 50404=1260密码

Example 3
::例3

How many ways can twelve students be organized evenly into classrooms A, B, C, and D?
::A、B、C和D教室的12名学生可以平均地组织多少种方式?

T here are twelve students, and four “teams” of three students each:
::共有12名学生,4个“小组”,每组3名学生:

$\begin{aligned} \frac{12!}{(3!) (3!) (3!) (3!)} \\ \frac{479001600}{1296} & = 369600 a r r a n g e m e n t s \end{aligned}$
$\begin{align*}& \frac{12!}{(3!)(3!)(3!)(3!)}\\ \frac{479001600}{1296}& =369600 \ arrangements\end{align*}$
::12号 12号 3号 3号 3号 47900016001296=369600安排

Review
::回顾

How many anagrams are possible using the letters in “possible”?
::使用这些字母的“可能”字句可以使用多少个方言?

How many anagrams are possible using the letters in “anagram”?
::使用“图表”中的字母可以使用多少个方程式?

Bobby knows he used the letters in his name, and the numbers in his age, 22, to make his e-mail password, mixing them up to make it hard to guess. Unfortunately, he made it too hard to guess, and forgot it. If the “reset password” option is unavailable, how many permutations might he need to guess?
::鲍比知道他用自己名字的字母和22岁时的数字来制作他的电子邮件密码,把密码混为一谈,使其难以猜测。不幸的是,他让自己太难猜,忘记了它。如果“重设密码”选项无法使用,那么他需要猜多少变数?

How many ways can fourteen students be seated, if one desk holds five students, one holds four students, one holds three, and one holds two?
::如果一个桌子容纳5名学生,一个桌子容纳4名学生,一个桌子容纳3名学生,一个桌子容纳2名学生,那么14名学生可以坐多少座位?

How many anagrams are possible using the letters in the word “letters”?
::使用“字母”一词中的字母可以使用多少个方字字?

There are thirteen pairs of shoes on the rack at a shoe store. There are four pairs of tennis shoes, three pairs of dress shoes, four pairs of high heels, and two pairs of sandals. How many ways can the pairs be arranged on the display shelf?
::鞋店的架子上有13双鞋,有四双网球鞋,三双洋装鞋,四双高跟鞋,两双凉鞋。在展品架上可以安排多少种方式?

There are five chocolate bars, three vanilla cookies, four chocolate chip cookies, and six snickerdoodle cookies on the counter. How many ways can the sweets be arranged on the counter?
::柜台上有5个巧克力棒,3个香草饼干,4个巧克力曲奇饼,6个饼干饼。在柜台上可以安排多少种甜点?

There are five apples, three bananas, four oranges, and three pears on a shelf. How many ways can they be arranged?
::5个苹果、3个香蕉、4个橙子和3个梨子放在架子上,可以安排多少种方式?

How many ways can thirteen friends be sorted into two groups of four and one group of five?
::13个朋友能分成两组 4个和5个?

How many ways can six nickels, five pennies, and four quarters be arranged?
::有多少方法可以安排六分五分和四分之四?

How many arrangements are possible using the numbers 1, 1, 3, 3, 3, 4, 5 and 5?
::使用数字1、1、3、3、3、3、4、4、5和5可以作出多少安排?

How many ways can three trucks, four motorcycles, four convertibles, and three SUV’s be parked all in a row?
::三辆卡车、四辆摩托车、四辆敞篷车和三辆SUV车能连续停靠多少方式?

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。

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