课程： 13.3 变换 | The Way To Learn

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Permutations
::差异

In this lesson we’ll be looking at ways of arranging things. To illustrate what we mean by this, let’s look at a simple example. Think about choosing your favorite color from the following list of choices; red, blue, green, yellow, pink, purple, orange, brown, black . Clearly there are nine different colors, so there are nine possible choices you could make.
::在这个教训中,我们将研究如何安排事情。为了说明我们的意思,让我们看看一个简单的例子。想一想从下面的选择列表中选择你最喜欢的颜色:红色、蓝色、绿色、黄色、粉红色、紫色、橙色、棕色、黑色。显然,有9种不同的颜色,所以你可以做出9种选择。

Now think about choosing your top three colors in order of preference. There are many different ways you can choose the top three. You might choose red as your favorite, followed by black, then green. Someone else might choose the same three colors as you, but in a different order! When you are choosing items from a list and the order in which you choose them is important , the arrangement is called a . How many different permutations do you think there are in this situation?
::现在考虑按偏好顺序选择您的前三个颜色。您可以选择前三个颜色。您可以选择许多不同的方式。您可以选择红色作为您的最爱, 之后是黑色, 然后是绿色。其他人可以选择与您相同的三个颜色, 但顺序不同。当您从列表中选择项目且您选择项目的顺序很重要时, 此安排被称为“ ” 。您认为在这种情况下会有多少不同的排列 ?

In this lesson, we’ll use counting methods to determine how many permutations a given situation has. We’ll also discover a formula to calculate permutations when counting alone is impractical.
::在这个教训中,我们将使用计数方法来确定特定情况有多少变异。我们还会发现一个公式来计算单独计时的变异是不切实际的。

Counting Permutations
::计算变数

In simple cases, sometimes it’s easiest to calculate permutations by just listing all the possibilities and counting them. Let’s examine a situation where it is relatively straightforward to do that.
::在简单的例子中,有时通过列出所有可能性并进行计算来计算变异有时比较容易。让我们来研究一个相对简单的情况。

1. Nadia and Peter are going to watch two movies on a rainy Saturday. Nadia will choose the first movie, and Peter gets to choose the second. The four movies they have to choose from are The Lion King, Aladdin, Toy Story and Pinocchio. Given that Peter will choose a different movie than Nadia, how many permutations are there for the movies they watch?
::1. Nadia和Peter将在雨季的星期六看两部电影,Nadia将选择第一部电影,Peter将选择第二部电影,他们必须选择的四部电影是狮子王、Aladdin、玩具童话和皮诺曹。鉴于Peter将选择与Nadia不同的电影,他们看的电影有多少变相?

Since the order in which they watch the movies is important, and they don’t plan to choose the same movie twice, we can list all the different possibilities in a table:
::因为他们看电影的顺序很重要, 他们不打算选择同一部电影两次, 我们可以在一张桌子上列出所有不同的可能性:

First Movie	Second Movie
Lion King	Aladdin
Lion King	Toy Story
Lion King	Pinocchio
Aladdin	Lion King
Aladdin	Toy Story
Aladdin	Pinocchio
Toy Story	Lion King
Toy Story	Aladdin
Toy Story	Pinocchio
Pinocchio	Lion King
Pinocchio	Aladdin
Pinocchio	Toy Story

You can see that this table contains all the possibilities for the situation. There are four movies for Nadia to choose from. For every movie that Nadia chooses first, Peter has three choices left for his movie. By simply counting the rows in the table you can see that there are 12 permutations in this situation.
::您可以看到这张表格包含所有可能性。 Nadia可以选择四部电影。对于Nadia首先选择的每部电影, Peter有三种选择可供他的电影使用。只需计算表格中的行数就可以看出, 在这种情形中,有12个变化。

2. I have 5 cards with the numbers 1 through 5 on them. I take three cards and arrange them to form a 3-digit number. How many 3-digit numbers can I make?
::2. 我有5张牌,上面有1至5号牌,我拿3张牌,并安排他们组成3位数的号码,我还能做多少个3位数的号码?

Since the numbers we can make fit a numerical ordering pattern, we can list the possibilities in increasing order:
::由于我们可以使数字符合数字顺序模式,我们可以列出增加顺序的可能性:

\begin{aligned} 123 124 125 132 134 135 142 143 145 152 153 154 \\ 213 214 215 231 234 235 241 243 245 251 253 254 \\ 312 314 315 321 324 325 341 342 345 351 352 354 \\ 412 413 415 421 423 425 431 432 435 451 452 453 \\ 512 513 514 521 523 524 531 532 534 541 542 543 \end{aligned}

By arranging the table this way, you can see how the number of remaining choices decreases as we choose numbers. There are five choices for the first number, four choices for the second number and three choices for the third number. Counting the table entries gives a total of 60 permutations .
::通过这样安排表格, 您可以看到剩余选择的数量会随着我们选择数字而减少。第一个数字有五个选择, 第二个数字有四个选择, 第三个数字有三个选择。计算表格条目时共显示60个变数。

If we look closely at the last two examples we can see a pattern start to appear. Mathematicians love patterns—they tend to lead to formulas, which make life much easier! After all, why spend hours counting possibilities when a formula can calculate them in seconds?
::如果我们仔细看看最后两个例子,我们可以看到一个模式开始出现。数学家的爱模式 — — 它们往往导致公式,让生活更加容易。毕竟,当公式能以秒计算时,为什么花几个小时来计算可能性呢?

In #1, Nadia had four choices and Peter had three. The number of permutations was $\begin{aligned} 4 \times 3 = 12. \end{aligned}$
::在1号中,Nadia有四个选择,Peter有三个选择。变换次数是4x3=12。

In #2 there were 5 choices for the first digit, followed by 4 for the second digit, and then 3 for the third digit. The total number of permutations was $\begin{aligned} 5 \times 4 \times 3 = 60 \end{aligned}$ .
::在 # 2 中, 首位数有 5 个选项, 第二位数后为 4 个, 第三位数后为 3 个。变换总数为 5x4x3= 60 。

In the introduction, we thought about picking our top three choices from a list of nine colors. You should now be able to do that. Even without listing all the possibilities, you can see that you have 9 choices for your favorite, 8 choices for your second favorite, and 7 choices for your third. The number of permutations is thus $\begin{aligned} 9 \times 8 \times 7 = 504 \end{aligned}$ .
::在引言中,我们考虑从九种颜色的列表中选择前三种选择。您现在应该能够这样做。即使没有列出所有的可能性, 您也可以看到您有9种选择, 8种选择第二最喜欢的, 7种选择第三种。因此, 变位数是 9x8x7=504 。

Factorial Notation
::因数标记

Look again at the color list in the introduction, and think this time about writing down every color in order of preference. You would have 9 choices for your favorite, followed by 8 choices for your second favorite, then 7, then 6, then 5, and so on. To determine the number of permutations for any possible list, we would perform the following calculation:
::再看看导言中的颜色列表, 并思考这次如何按偏好顺序写下每个颜色。您的最爱有9种选择, 其次是8种选择, 其次是第二最爱, 然后是7, 然后是6, 然后是5, 等等。为了确定任何可能的列表的变异数量, 我们将进行以下计算 :

$\begin{aligned} Color Permutations = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \end{aligned}$
::颜色间距=9x8x7x6x5x4x4x3x3x2x1

This sort of pattern crops up a great deal in statistics, probability and number theory. It’s so common that it has its own notation: $\begin{aligned} 4 \times 3 \times 2 \times 1 \end{aligned}$ is written as 4! and is called four . So the number of color permutations above is nine factorial = 9! = 362,880.
::这种类型的模式作物在统计、概率和数字理论方面有很大的上升。它非常常见, 以至于它有自己的符号 : 4x3x2xx1 写为 4! , 叫做 4 。上面的颜色变异数是 9 阶乘= 9! = 362, 880 。

So what happens when we only want the first few terms in a factorial? For example, the number of permutations for arranging ALL the colors is 362,880, but the number of permutation for the top three is 504.
::那么当我们只想要一个因素的最初几个条件时会发生什么呢?例如,排列所有颜色的变换数量是362,880,但前三个变换数量是504。

One way to get this result is to divide one factorial by another. Look at nine factorial divided by six factorial:
::取得这一结果的一个办法是将一个因数除以另一个。看看九个因数除以六个因数:

\begin{aligned} \frac{9!}{6!} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 9 \times 8 \times 7 = 504 \end{aligned}

The terms in six factorial cancel out all but the first three terms in nine factorial. You should see that if we wanted the first four terms we would divide by 5! , or for only the first two terms we would divide by 7! . In general, however many terms we want to keep, we divide by the factorial of the quantity:
::6个因数的术语除前3个因数的9个因数的术语外,除前3个因数的术语外,全部取消。你应该看到,如果我们想要前4个术语,我们就会除以5! 或者仅仅前2个术语,我们就会除以7! 一般来说,无论我们想保留多少术语,我们都要以数量因数的因数来划分:

\begin{aligned} (n u m b e r o f i t e m s i n l i s t) - (n u m b e r o f i t e m s w e a r e c h o o s i n g) \end{aligned}

清单中项目数目)-(我们所选择的项目数目)

So to get the first five terms in twelve factorial we would use the formula $\begin{aligned} \frac{12!}{(12 - 5)!} = \frac{12!}{7!} \end{aligned}$ .
::因此,为了在12个阶乘中获得前5个条件,我们将使用12号公式! (12) - 5!=12! 7!

Formulas like this are useful if you have a calculator that can handle factorials: you can just type in $\begin{aligned} \frac{12!}{7!} \end{aligned}$ instead of $\begin{aligned} 12 \times 11 \times 10 \times 9 \times 8 \end{aligned}$ . However, some factorials are too big for some calculators to handle, so in those cases you would need to simplify the fraction and do the multiplication by hand.
::像这样的公式有用, 如果您有计算器可以处理阶乘数: 您只需输入 12! 7! 而不是 12 x11 x10 x9 x8. 。然而, 有些阶乘数太大, 某些计算器无法处理, 所以在这样的情况下, 您需要简化分数, 并用手进行乘法。

Using Factorials
::使用因数

How many ways can Dale choose his favorite 5 songs from the current Billboard Hot $\begin{aligned} 100^{T M} \end{aligned}$ ?
::戴尔能用多少方法选择他最喜欢的五首歌从目前的广告牌热100TM?

To find the answer, consider how many choices he has at each stage. For his first choice he has 100 songs to choose from, then 99, then 98 and so on. We need the first 5 terms only, so our calculation is:
::为了找到答案,请考虑一下他在每个阶段有多少选择。他的首选有100首歌曲,然后是99首,然后是98首等等。我们只需要前5个条件,所以我们的计算是:

\begin{aligned} Permutations = \frac{100!}{(100 - 5)!} = \frac{100!}{95!} & = 100 \times 99 \times 98 \times 97 \times 96 \\ = 9, 034, 502, 400 \end{aligned}

::变异=100!(100-5)! =100!95! =100x99x98x98x97x96=9,034,502,400

Notice that that’s a pretty big number – far too large to count in a table! This is why we need formulas to help us count permutations.
::请注意这是一个相当大的数字 — — 太大到无法在桌子上计数! 这就是为什么我们需要公式来帮助我们计算变数。

Finding Permutations Using a Formula
::使用公式查找差异

We’ve just seen that a formula for determining the number of permutations for choosing 3 objects from a list of 9 objects is:
::我们刚刚看到,从9个天体的列表中选择3个天体,确定变异数目的公式是:

\begin{aligned} \frac{9!}{6!} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 9 \times 8 \times 7 = 504 permutations \end{aligned}

::9! 6! =9x8x7x6x5x5x4x3x3x2x16x5x4x3x2x1=9x8x7=504

Now we’re ready to come up with a general formula for determining permutations. When we are choosing $\begin{aligned} r \end{aligned}$ ordered items from a group of $\begin{aligned} n \end{aligned}$ items, the number of permutations is given by the first $\begin{aligned} r \end{aligned}$ terms in $\begin{aligned} n! \end{aligned}$ We use the notation $\begin{aligned} _{n} P_{r} \end{aligned}$ for this, and the general form for calculating permutations is:
::现在,我们准备提出一个用于确定变价的一般公式。当我们从一组 n 项中选择 r 订购的项目时, 变价数由n 中的第一个 r 条件给出。我们为此使用nPr 标记, 计算变价的一般形式是 :

\begin{aligned} _{n} P_{r} = \frac{n!}{(n - r)!} \end{aligned}

::NPRN!

Calculating Permutations
::计算间替

How many ways are there to choose a 5-song mix from a CD containing 12 tracks?
::从含有12个音轨的CD中选择五方位组合有多少方法?

Choosing 5 from 12: $\begin{aligned} _{12} P_{5} = \frac{12!}{(12 - 5)!} = \frac{12!}{7!} = 12 \times 11 \times 10 \times 9 \times 8 = 95, 040 w a y s \end{aligned}$
::从 12: 12P5=12 选择 5! (12- 5)!= 12! 7!= 12x11x10x10x9x8=95, 040 方式

Example
::示例示例示例示例

Example 1
::例1

How many 3-letter “words” can be made from the letters in “computer”? (The words do NOT need to be real, or even pronounceable – for example “rtp” would count as a word)
::从“计算机”中的字母中可以写出多少3个字母的“字”? (这些字不必是真实的,甚至不必是可宣布的——例如,“rpp”将算作一个字)

Choosing 3 from 8: $\begin{aligned} _{8} P_{3} = \frac{8!}{(8 - 3)!} = \frac{8!}{5!} = 8 \times 7 \times 6 = 336 w o r d s \end{aligned}$
::从 8: 8 8P3= 8! (8- 3)! = 8! 5! = 8x7x6= 336 单词

Review
::回顾

In how many ways can the letters $\begin{aligned} a, b, c, d, e \end{aligned}$ be arranged?
::可以用多少方式安排字母 a,b,c,d,e?

In how many ways can the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 be arranged?
::数字1、2、3、4、5、6、7、8、9可以安排多少种方式?

From a collection of 12 books, 5 are to be selected and placed in a particular order on a shelf. How many arrangements are possible?
::从12本书集中挑选5本,按特定顺序放在一个架子上,可以作出多少安排?

3 cards are taken at random from a deck of 52 cards and laid in a row. How many possible outcomes are there for the card arrangements?
::从52张牌牌牌的甲板上随机抽取3张牌,并排成一排。

How many distinct 3-letter permutations can you make from the letters in the word HEXAGON?
::你能用HEXAGON字里的字母做多少个不同的3字母排列?

How many distinct 2-letter permutations can you make from the letters in the word GEESE?
::你能用 GEESE 这个词里的信件做多少个不同的两个字母的排列?

A jukebox has 50 songs on it. If $1.00 pays for three songs, how many permutations are there for choosing 3 different songs?
::一个点唱机上面有50首歌。如果1美元支付3首歌,那么选择3首不同的歌曲有多少配音?

For problems 8-16, evaluate the following:
::对于问题8-16,评估如下:

$\begin{aligned} _{3} P_{1} \end{aligned}$
::3P1 3P1

$\begin{aligned} _{7} P_{1} \end{aligned}$
::7P1 7P1

$\begin{aligned} _{6} P_{2} \end{aligned}$
::6P2 6P2

$\begin{aligned} _{8} P_{8} \end{aligned}$
::8P88

$\begin{aligned} _{9} P_{3} \end{aligned}$
::9P3

$\begin{aligned} _{7} P_{3} \end{aligned}$
::7P3 7P3

$\begin{aligned} _{19} P_{7} \end{aligned}$
::19P7

$\begin{aligned} _{99} P_{3} \end{aligned}$
::99P3

$\begin{aligned} _{3} P_{0} \end{aligned}$
::3P0 3P0

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

常规

差异

Permutations ::差异

Counting Permutations ::计算变数

Factorial Notation ::因数标记

Using Factorials ::使用因数

Finding Permutations Using a Formula ::使用公式查找差异

Calculating Permutations ::计算间替

Example ::示例示例示例示例

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)