Course: 12.13 变异问题 | The Way To Learn

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变异问题
Veronica is a lifeguard at the community swimming pool. She is in charge of creating an order for the 10 children who are waiting to use the diving board. She can send them four at a time to line up. How many different combinations can Veronica make?
::维罗妮卡是社区游泳池的救生员,她负责为10名正在等待使用潜水板的儿童制定命令。她可以一次派4名儿童排队。维罗妮卡能做多少不同的组合?

In this concept, you will learn about counting permutations.
::在这个概念中,你会学会计算变数。

Counting Permutations
::计算变数

Once you decide that order does matter, you know that you are working with a . It is helpful to know how to calculate permutations.
::一旦您决定该顺序很重要, 您就会知道您正在与 . 合作。了解如何计算变数会很有帮助。

Let’s look at an example to see how to do this.
::让我们举个例子看看如何做到这一点。

The softball coach needs to determine how many different batting lineups she can make out of her first three batters, Able, Baker, and Chan. How many different batting orders are there?
::垒球教练需要确定她能从前三个击球手Able、Baker和Chan身上得出多少不同的击球队。有多少不同的击球命令?

One way to look at this problem is as the product of 3 different choices. For choice 1 you can select any of the three batters, Able, Baker, or Chan.
::看待这个问题的一种方式是作为三个不同选择的产物。对于选择 1 , 您可以选择三个击球手中的任何一个, Able 、 Baker 或 Chan 。

$\begin{aligned} Choice 1 \times Choice 2 \times Choice 3 = Possible Number of Choices \end{aligned}$

::选择 1x 选择 2x 选择 3 = 可能的选择数

3 is Choice 1 because you have 3 batters to choose from.
::3是选择1,因为您有3个击球手可供选择。

2 is Choice 2 because you selected one batter for Choice 1 leaving 2 choices.
::2是选择2,因为您选择了 1 选择 1 选择 2 选择。

1 is Choice 3 because that is all that is left.
::选择31就是选择3,因为剩下的就只有这个了。

The answer is $\begin{aligned} 3 \times 2 \times 1 = 6 \end{aligned}$ . There are six possible orders.
::答案是 3x2x1=6 有六个可能的命令

Using the Counting Principle you can multiply the three choices together to get the total number of choices, or permutations, as 6. Here are the 6 different batting lineups.
::使用计数原则,您可以将三种选择一起乘以,以获得选择的总数,或变换,如6。这是6个不同的击球队。

$\begin{aligned} Able-Baker-Chan Baker-Able-Chan Chan-Able-Baker \\ Able-Chan-Baker Baker-Chan-Able Chan-Baker-Able \end{aligned}$

::易易易易易易易易易易易易易易

Notice that order is important here. Each of the 6 choices, or permutations, is a unique and different batting order. For example, Able-Baker-Chan is not the same lineup as Able-Chan-Baker.
::注意命令在这里很重要。 6种选择, 或变换, 每一个都是独特的和不同的打击顺序。例如, Able- Baker- Chhan 和 Able- Chander 不一样。

What happens when you increase the number of players in your lineups by adding Davis? How many different lineups are there now? Starting over, you can now see that there are 4 choices for the first batter, followed by 3 choices, 2 choices, and 1 choice.
::当你通过添加戴维斯来增加队列中的球员人数时会怎样?现在有多少队列?从现在开始,你可以看到第一击手有4个选择,接着是3个选择,2个选择,1个选择。

$\begin{aligned} 4 \times 3 \times 2 \times 1 = 24 possible choices \end{aligned}$

::4x3x2x1=24 可能的选项

Let’s look at another example.
::让我们再看看另一个例子。

How many different arrangements of the letters $\begin{aligned} A, B, C, D \end{aligned}$ , and $\begin{aligned} E \end{aligned}$ can you make without repeating any of the letters?
::A、B、C、D和E字母的多少不同安排,

This is very much like the Able, Baker, Chan, Davis problem only it adds a fifth item. Notice how this extra item increases the total by a huge amount. In fact, there are exactly 5 times as many permutations of 5 items than there were of 4 items above.
::这与Able、Baker、Chan、Davis问题非常相似,它只增加了第五个问题。请注意这一额外项目是如何使总数增加巨大数额的。事实上,5个项目的变位数是上面4个项目的5倍。

$\begin{aligned} 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120 \\ choice 1 choice 2 choice 3 choice 4 choice 5 total choices \end{aligned}$

::54321=120选1 选2 选3 选3 选4 选5 总选5

You learned how to count permutations of a certain amount of choices where all of the choices were used each time. For instance, when you completed the batter line up of three batters in order, you could count the permutations like this.
::您学会了如何计算每次所有选择都使用的一定数量的选择的变异。例如,当你完成击打线时, 三个击打者排成一排, 您可以计算这样的变异。

$\begin{aligned} 3 \times 2 \times 1 = 6 possible permutations \end{aligned}$

::3x2x1=6 可能的变换

What happens if you have four players, but you only wanted to put three in the line up? This changes the way that you count permutations. To accomplish this task, you start counting at 4 and then find the product of the next two values as well.
::如果你有四个玩家,但你只想在队列中加上三个,会怎么样?这改变了你计算变数的方式。要完成这项任务,你开始数到4,然后也找到下两个数值的产物。

$\begin{aligned} 4 \times 3 \times 2 = 24 permutations \end{aligned}$

::4x3x2=24 变换

Notice that you don’t include the 1 because there are four batters taken three at a time.
::请注意您不包含1, 因为有4名殴打者一次3人。

Let’s look at another example.
::让我们再看看另一个例子。

What will happen if Elvis joins the team? Now, with 5 players to choose from, how many different 3-player batting orders are there?
::如果猫王加入球队怎么办?

This time you have 5 different players for your first choice, 4 players for your second choice, and 3 players for your third choice.
::这次你有5个不同的选手 4个选手做第二选 3个选手做第三选

$\begin{aligned} 5 \times 4 \times 3 = 60 options \end{aligned}$

::5x4x3=60 选项

Taken 4 at a time, how many different arrangements of the letters $\begin{aligned} A, B, C, D, E \end{aligned}$ , and $\begin{aligned} F \end{aligned}$ are there?
::A、B、C、D、E和F字母有几处不同的安排?

First, notice that there are six possible letter choices to work with. You want to take them four at a time.
::首先,请注意,有六个可能的字母选择可以一起工作。你要一次用四个字母。

$\begin{aligned} 6 \times 5 \times 4 \times 3 = 360 possible options \end{aligned}$

::6x5x4x3=360 可能的选项

You just figured out permutations by arranging numbers. You can use permutation notation to help you know when you are working with a permutation. Permutation notation involves something called a . A factorial is a way of writing a number to show that you are going to be looking for the product of a series of numbers. The symbol for a factorial is an exclamation sign.
::您只是通过安排数字来理解变异。您可以使用变异符号来帮助您知道您在使用变异。变异符号包含一个叫做。乘数是一种写数字的方式, 以显示您将寻找一系列数字的产物。乘数的符号是感叹符号。

Here are some factorials.
::以下是一些因素。

$\begin{aligned} 8! & = 8 - factorial \\ 11! & = 11 - factorial \\ 29! & = 29 - factorial \\ 2! & = 2 - factorial, and so on \end{aligned}$

::8! =8 - 省11! =11 - 省29! =29 - 省2! =2 - 省2,等等

To compute factorials, rewrite the number before the exclamation point and all of the whole numbers that come before it.
::要计算阶乘,请在感叹点之前重写数字,并重写全部数字。

$\begin{aligned} 4! & = 4 \cdot 3 \cdot 2 \cdot 1 \\ 7! & = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \\ 2! & = 2 \cdot 1 \\ 11! & = 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \end{aligned}$

What are the values of these factorial numbers?
::这些系数数字的值是什么?

$\begin{aligned} 4! & = 4 \cdot 3 \cdot 2 \cdot 1 = 24 \\ 5! & = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120 \\ 7! & = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5040 \end{aligned}$

That is all that you need to know. As long as you remember how to find the product of a factorial, you will always know a short cut for permutations!
::这就是你需要知道的。只要你记得如何找到一个阶乘的产物, 你就会永远知道一个短距离的变换!

Now that you’ve learned how to work with factorials, you are ready see how to use them to calculate permutations. Suppose you have 5 letters – $\begin{aligned} A, B, C, D \end{aligned}$ , and $\begin{aligned} E \end{aligned}$ . You want to know how many permutations there are if you take 3 letters at a time and find all arrangements of them. In permutation notation you write this as:
::既然你已经学会了如何与阶乘法打交道,你就可以看到如何使用它们来计算变数。如果你有5个字母 — — A、B、C、D和E。你想知道如果每次用3个字母并找到所有字母的排列方法,其中会有多少变数。在变位符号中,你写为:

$\begin{aligned} _{5} P_{3} ⟸ \end{aligned}$ 5 items taken 3 at a time
::5P3 5件物品一次3件

In general, permutations are written as:
::一般说来,变式的写法是:

$\begin{aligned} _{n} P_{r} ⟸ n \end{aligned}$ items taken $\begin{aligned} r \end{aligned}$ at a time
::nPrn 项目一次拍摄r

To compute $\begin{aligned} _{n} P_{r} \end{aligned}$ you write:
::要计算 nPr, 您要写入 :

$\begin{aligned} _{n} P_{r} = \frac{n!}{(n - r)!} = \frac{⟸ total items!}{⟸ (total items-items taken at a time)!} \end{aligned}$

::nPr=n! (n-r)! (jects-items- jects at a time)! (jects- jects- jects at a time) ! (n-r)! (n-r)!

To compute $\begin{aligned} _{5} P_{3} \end{aligned}$ just fill in the numbers:
::要计算 5P3 只填入数字:

$\begin{aligned} _{5} P_{3} & = \frac{5!}{(5 - 3)!} = \frac{⟸ total items!}{⟸ (total items-items taken at a time)!} \\ _{5} P_{3} & = \frac{5 (4) (3) (2) (1)}{2 (1)} = \frac{120}{2} = 60 \end{aligned}$

::5P3=5! 5P3=5! (5- 3)! * 总计项目! * * (一次使用的项目总数)! 5P3=5(4)(3)(2)(1)(2)(1)(1)=1202=60

There are 60 possible permutations.
::有60种可能的变异。

You can use this formula anytime that you are looking to figure out permutations!
::您可以随时使用此公式, 以便找出变数 !

Examples
::实例

Example 1
::例1

Earlier, you were given a problem about Veronica's responsibility of sending four children to line up at the diving board at a time.
::早些时候,你被问及维罗妮卡的责任送四个孩子去潜水板排队

There are a total of 10 children. How many different combinations of children can Veronica make?
::共有10个孩子维罗妮卡能做多少种不同的组合?

First, substitute the values into the appropriate place in the formula.
::首先,将数值改为公式中的适当位置。

$\begin{aligned} _{1} 0 P_{4} & = \frac{10!}{(10 - 4)!} = \frac{⟸ total items!}{⟸ (total items-items taken at a time)!} \\ _{1} 0 P_{4} & = \frac{(10) (9) (8) (7) (6) (5) (4) (3) (2) (1)}{(6) (5) (4) (3) (2) (1)} \end{aligned}$
Next, multiply all values together.
$\begin{aligned} _{1} 0 P_{4} & = \frac{(10) (9) (8) (7) (6) (5) (4) (3) (2) (1)}{(6) (5) (4) (3) (2) (1)} = \frac{3, 628, 800}{720} \end{aligned}$
Then, divide the numerator by the denominator.
$\begin{aligned} _{1} 0 P_{4} & = \frac{3, 628, 800}{720} = 5, 040 \end{aligned}$
The answer is there 5,040 possible combinations.
::10P4=10!(10-4(4) = 10); = 总计项目! = (一次使用的项目总数)! 10P4=(10)(9)(9)(7)(6)(6)(6)(6)(4)(4)(3)(2)(1)下一步,所有数值乘以10(6)(4)(4)(10)(9)(8)(7)(6)(4)(4)(4)(3)(3)(2)(2)(2)(2)(2)(2)=3,628800720 然后将分子除以分母10.4=3,628 800720=5,040 答案是5,040可能的组合。

Example 2
::例2

Calculate the following permutation:
::计算以下排列:

$\begin{aligned} _{9} P_{5} \end{aligned}$
::9P5

First, substitute the values into the appropriate place in the formula:
::首先,将数值改为公式中的适当位置:

$\begin{aligned} _{9} P_{5} & = \frac{9!}{(9 - 5)!} = \frac{⟸ total items!}{⟸ (total items-items taken at a time)!} \\ _{9} P_{5} & = \frac{9 (8) (7) (6) (5) (4) (3) (2) (1)}{(4) (3) (2) (1)} \end{aligned}$

::9P5=9!9P5=9!(9-5)!* 总计项目!* (一次使用的项目总数)!9P5=9(8)(7)(6)(5)(4)(4)(3)(3)(2)(1)(4)(4)(3)(1)(1)(1)

Next, multiply all values together:
::下一步,将所有值相乘:

$\begin{aligned} _{9} P_{5} & = \frac{9 (8) (7) (6) (5) (4) (3) (2) (1)}{(4) (3) (2) (1)} = \frac{362880}{24} \end{aligned}$

::9P5=9(8)(7)(6)(5)(4)(3)(2)(2)(1)(4)(3)(2)(2)(1)=36288024

Then, divide the numerator by the denominator:
::然后,将分子除以分母:

$\begin{aligned} _{9} P_{5} & = \frac{362880}{24} = 15, 120 \end{aligned}$
The answer is there are 15,120 possible combinations.
::9P5=36288024=15120 答案是15,120个可能的组合。

Calculate the following permutations.
::计算以下的变差。

Example 3
::例3

$\begin{aligned} _{3} P_{2} \end{aligned}$
::3P2 3P2

First, substitute the values into the appropriate place in the formula:
$\begin{aligned} _{3} P_{2} & = \frac{3!}{(3 - 2)!} = \frac{⟸ total items!}{⟸ (total items-items taken at a time)!} \\ _{3} P_{2} & = \frac{(3) (2) (1)}{(1)} \end{aligned}$
Next, multiply all values together:
$\begin{aligned} _{3} P_{2} & = \frac{(3) (2) (1)}{(1)} = \frac{6}{1} \end{aligned}$
Then, divide the numerator by the denominator:
$\begin{aligned} _{3} P_{2} & = \frac{6}{1} = 6 \end{aligned}$
The answer is there are 6 possible combinations.
::首先,将值替换为公式3P2=3!(3-2)!{{{{{{}}}}}{}}}{}}}{}}}{}}}}{}}}}}{}}}}{}}}}}{}}}}{}}}}{}}}}{}}}{}}}}}}}{}}}}}}}}{}}}}}{}}}}}}{}}}}{}}}}}}}{}}}}}}{}}}}}{}}}}{}}}}}}{}}}}}}{}}}}}{}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}。}}}}}}{}}}}}}}}}}}}}}}}}}}}}。}}}}}}}}。}}}}}。}}}}}}}}}}}}}}}}}}}}}}}}。}。}}}}}}}}}。}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}。}}}}}}}}}}}}}}}}}}}}。}}}}}}}}}}}}}}}}}}}}}}}。}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。

Example 4
::例4

$\begin{aligned} _{5} P_{4} \end{aligned}$
::5P4 5P4

First, substitute the values into the appropriate place in the formula:
$\begin{aligned} _{5} P_{4} & = \frac{5!}{(5 - 4)!} = \frac{⟸ total items!}{⟸ (total items-items taken at a time)!} \\ _{5} P_{4} & = \frac{(5) (4) (3) (2) (1)}{(1)} \end{aligned}$
Next, multiply all values together:
$\begin{aligned} _{5} P_{4} & = \frac{(5) (4) (3) (2) (1)}{(1)} = \frac{120}{1} \end{aligned}$
Then, divide the numerator by the denominator:
$\begin{aligned} _{5} P_{4} & = \frac{120}{1} = 120 \end{aligned}$
The answer is there are 120 possible combinations.
::首先,将值替换为公式中的适当位置:5P4=5!!(5-4)!{{{{{{{}}}}}{}}{}}}{}}}{}}}}{}}}}{}}}}{}}}}{}}}}{}}}}{}}}}{}}}}}}}{}}}}}}}{}}}}}{}}}}}}}{}}}}{}}}}}}}}}{}}}}}{}}}}{}}}}}{}}}}}}{}}}}{}}}}}}{}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}{}}}}}}}}}}}}{}}}}}}}}}}}}}}}}}}{}}}}}}}}}}}}}}}}。}}}}。}}}}}}}}}}}。}}}}}}}}}}}}}}}}}}}}}}}}}}}}}。}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}。}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}。}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}

Example 5
::例5

$\begin{aligned} _{9} P_{2} \end{aligned}$
::9P2

First, substitute the values into the appropriate place in the formula:
$\begin{aligned} _{9} P_{2} & = \frac{9!}{(9 - 2)!} = \frac{⟸ total items!}{⟸ (total items-items taken at a time)!} \\ _{9} P_{2} & = \frac{9 (8) (7) (6) (5) (4) (3) (2) (1)}{(7) (6) (5) (4) (3) (2) (1)} \end{aligned}$
Next, multiply all values together:
$\begin{aligned} _{9} P_{2} & = \frac{9 (8) (7) (6) (5) (4) (3) (2) (1)}{(7) (6) (5) (4) (3) (2) (1)} = \frac{362, 880}{5, 040} \end{aligned}$
Then, divide the numerator by the denominator:
$\begin{aligned} _{9} P_{2} & = \frac{362880}{5, 040} = 72 \end{aligned}$
The answer is there are 72 possible combinations.
::首先,将数值改为公式9P2=9!(9-2)!(总项数)!(一次使用的项目总数)!9P2=9(8)(7)(6)(4)(3)(2)(7)(6)(5)(4)(3)(3)(2)(1)下一步,将所有值乘以:9P2=9(8)(6)(6)(4)(4)(3)(2)(7)(7)(6)(6)(6)(5)(4)(3)(4)(3)(2)(6)(6)(5)(4)(3)(2)=3628805,040Text,将分子除以分母:9P2=3628805.040=72。答案是:72个可能的组合。

Review
::回顾

Count all of the permutations.
::计数所有变换。

Three marbles–red, blue, yellow,–are in a jar. In how many different orders can you pull two of the marbles out of the jar (without replacing either of the marbles in the jar)?
::三个大理石 — — 红色、蓝色、黄色 — — 放在一个罐子里。你能用多少种不同的命令把两个大理石从罐子里拉出来(而不是取代罐子里的任何一个大理石 ) ?

A green marble is added to the jar above, giving red, blue, yellow and green marbles. In how many different orders can you pull three of the marbles out of the jar (without replacing any of the marbles in the jar)?
::绿色大理石被添加到上面的罐子上,给出红色、蓝色、黄色和绿色大理石。您可以用多少种不同的命令从罐子上拉出三个大理石( 而不替换罐子上的任何大理石 ) ?

In a jar with 4 marbles–red, blue, yellow, and green–how many different orders will you have if you pull just 2 marbles from the jar (without replacing either of the marbles in the jar)?
::在一个装有4大理石的罐子里, 红色、蓝色、黄色和绿色, 如果你从罐子上拿出2大理石(而不替换罐子中的任何一个大理石), 你会有多少不同的命令?

In a jar with 5 marbles–red, blue, yellow, green, and white–how many different orders will you have if you pull 3 marbles from the jar?
::在一个有五大理石的罐子里, 红色、蓝色、黄色、绿色和白色, 如果你从罐子上拔出三大理石, 你会有多少不同的订单?

How many 4-digit arrangements can you make of the digits 1, 2, 3, 4, 5 if you don’t repeat any digit?
::如果您不重复任何数字, 数字1、 2、 3、 4、 5, 您可以做出多少四位数的安排?

A TV channel has 6 different 1-hour shows to fill 3 hours of time for Thursday night. How many different program lineups can the channel present?
::电视频道有6个不同的1小时节目,用来填充星期四晚上的3小时时间。频道可以播放多少个不同的节目排队?

Seven ski racers compete in the finals of the slalom event. In how many different orders can the top 3 skiers finish?
::七个滑雪赛车手在滑雪比赛的决赛中竞争。排在前三位的滑雪运动员能完成多少不同的订单?

Seven ski racers compete in the finals of the downhill event. In how many different orders can the top 4 skiers finish?
::7名滑雪赛车手在下坡赛决赛中竞争。前4名滑雪运动员能完成多少不同的订单?

Solve each factorial.
::解决每个因素。

5!

3!

6!

4!

Use permutation notation and the formula to find each permutation.
::使用变异符号和公式来查找每种变异。

$\begin{aligned} _{4} P_{2} \end{aligned}$
::4P2 4P2

$\begin{aligned} _{4} P_{3} \end{aligned}$
::4P3 4P3

$\begin{aligned} _{5} P_{4} \end{aligned}$
::5P4 5P4

$\begin{aligned} _{6} P_{3} \end{aligned}$
::6P3 6P3

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。

Resources
::资源

Section outline

General

变异问题

Counting Permutations ::计算变数

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)

Resources ::资源

Counting Permutations
::计算变数

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Review
::回顾

Review (Answers)
::回顾(答复)

Resources
::资源