Course: 12.19排列 | The Way To Learn

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排列

Coach Coker is focused on building the defensive team. During games, the defensive team will need two players in the safety position to guard against long passes. To make sure the team has enough players who can play safety, Coach Coker wants to know how many different combinations of safety position players the team has if he has 6 players on the team who can play safety. The order of the arrangements is important because some of the players play other positions. How many combinations or permutations of two players are possible if order is important?
::库克教练(Coker教练)专注于建立防御小组。在游戏中,防御小组需要两个处于安全位置的球员来防范过长的球员。为了确保团队有足够的球员来玩安全游戏,科克教练想知道,如果团队中有6个球员可以玩安全游戏,小组有多少个不同的安全位置球员组合。这些安排的顺序很重要,因为有些球员在玩其他位置。如果秩序重要的话,两个球员有多少组合或组合是可能的?

In this concept, you will learn about permutations and how to find all possible permutations.
::在这个概念中,你会了解变异和如何找到所有可能的变异。

Finding Permutations
::查找差异

A is a where order makes a difference.
::A是秩序改变的地方。

Look at the combinations listed in the chart below.
::看看下面图表中列出的组合。

SK	KS	DS	JS
SD	KD	DK	JK
SJ	KJ	DJ	JD

There are 12 possible outcomes for this permutation because order is important and you do not to get rid of duplicates.
::这种变换有12种可能的结果,因为秩序很重要,不能消除重复。

Using specific notation is an easier way to figure permutations than writing out all of the possibilities.
::使用具体标记比写出所有可能性更容易显示变异。

For example, in this chart there are initials for four boys in pairs - four taken two at a time. Here is how to write this as a permutation.
::例如,在这张图表中,有4个男孩一对的首字母缩写—— 4个男孩一次2个。下面是如何写成一个变体。

$\begin{aligned} P (4, 2) \end{aligned}$
::P(4 4 , 2 )

This means there are four options taken two at a time.
::这意味着有四个选项, 一次使用两个选项。

To figure out the permutation, count down from 4 two numbers (4 and 3) then multiply them. There are two numbers to multiply because the boys were arranged two at a time.
::要找出变异情况, 从 4 2 个数字( 4 和 3) 中倒计数, 然后乘以 4 个数字( 4 和 3 ) 。有 2 个数字要乘以。因为男孩子是一次安排两个数字。

\begin{aligned} 4 \times 3 = 12 \end{aligned}

There are 12 possible combinations. That is the same answer you found by writing things all out.
::有12个可能的组合。这是相同的答案你通过写出所有的东西找到的答案。

Here is another example.
::下面是另一个例子。

How many ways can you arrange five swimmers in groups of three?
::你有多少方法可以安排五个游泳者一组三人?

First, because you have groups of 3, multiply together the last 3 numbers in the count up to the number of items. Here is the permutation of 5 taken three at a time.
::首先,因为您有 3 组, 将最后 3 个数字乘以 3 个数字, 最多到数项。这是 5 的变换, 一次三次。

\begin{aligned} P (5, 3) = 5 \times 4 \times 3 \end{aligned}

There are 60 possible combinations.
::有60种可能的组合。

Examples
::实例

Example 1
::例1

Earlier, you were given a problem about Coach Coker and his defensive team.
::早些时候,你遇到了一个科克教练和他的防御队的问题

How many different combinations of 6 players in the 2 safety positions are there? When it comes to pairing the two players in the position, the order of the pairings is important.
::在2个安全位置上,6个球员的组合有多少?在对两个球员进行配对时,配对的顺序很重要。

First, express this as $\begin{aligned} P (6, 2) \end{aligned}$ , which means there are 5 players taken 2 at a time.
::首先,以P(6,2)表示这一点,这意味着有5个球员,每次2个球员。

$\begin{aligned} P (6, 2) \end{aligned}$
:P 6, 2 )

Next, note the last two numbers leading up to 6. You use two because there are 2 to each pairing.
::接下来,注意最后两个数字导致到6。你使用两个,因为每个配对有2对。

6, 5

Then, multiply the two numbers to get the permutations.
::然后,乘以这两个数字以获得变换。

\begin{aligned} \begin{array}{rcl} P (6, 2) & = & 6 \times 5 \\ P (6, 2) & = & 30 \end{array} \end{aligned}

The answer is 30.
::答案是30岁

There are 30 permutations.
::有30种变异。

Example 2
::例2

Evaluate $\begin{aligned} P (8, 3) \end{aligned}$ .
::评价P(8,3)

First, note the last three numbers leading up to 8. You use three because there are 3 to each group.
::首先,注意最后三个数字导致到8。你使用三个,因为每组有3个。

8, 7, 6

Next, multiply the three numbers to get the permutations.
::下一个,乘以三个数字以获得变换。

\begin{aligned} \begin{array}{rcl} P (8, 3) & = & 8 \times 7 \times 6 \\ P (8, 3) & = & 336 \end{array} \end{aligned}

The answer is 336.
::答案是336

Example 3
::例3

Evaluate $\begin{aligned} P (9, 2) \end{aligned}$ .
::评价P(9,2)

First, note the last two numbers leading up to 9. You use two because there are 2 to each group.
::首先,注意最后两个数字导致到9。你使用两个,因为每组有2个。

9, 8

Next, multiply the numbers to get the permutations.
::下一个,乘以数字来获得变换。

\begin{aligned} \begin{array}{rcl} P (9, 2) & = & 9 \times 8 \\ P (9, 2) & = & 72 \end{array} \end{aligned}

The answer is 72.
::答案是72。

Example 4
::例4

Evaluate $\begin{aligned} P (4, 3) \end{aligned}$ .
::评价P(4、3)

First, note the last three numbers leading up to 4. You use three because there are 3 to each group.
::首先,注意最后三个数字导致4。你使用三个,因为每组有3个。

4, 3, 2

Next, multiply the three numbers to get the permutations.
::下一个,乘以三个数字以获得变换。

\begin{aligned} \begin{array}{rcl} P (4, 3) & = & 4 \times 3 \times 2 \\ P (4, 3) & = & 24 \end{array} \end{aligned}

The answer is 24.
::答案是24个。

Example 5
::例5

Evaluate $\begin{aligned} P (5, 2) \end{aligned}$ .
::评价P(5,2)

First, note the last two numbers leading up to 5. You use two because there are 2 to each group.
::首先,注意最后两个数字,直到5。你使用两个,因为每个组有2个。

5, 4

Next, multiply the two numbers to get the permutations.
::下一个,乘以这两个数字以获得变换。

\begin{aligned} \begin{array}{rcl} P (5, 2) & = & 5 \times 4 \\ P (5, 2) & = & 20 \end{array} \end{aligned}

The answer is 20.

::答案是20岁

Review
::回顾

Determine the following permutations.
::确定下列变化。

$\begin{aligned} P (5, 2) \end{aligned}$
:P 5, 2 )

$\begin{aligned} P (6, 3) \end{aligned}$
::P(6 6 , 3 )

$\begin{aligned} P (7, 2) \end{aligned}$
:P 7, 2 )

$\begin{aligned} P (5, 4) \end{aligned}$
:P 5 , 4 )

$\begin{aligned} P (7, 3) \end{aligned}$
::P( 7 , 3 )

$\begin{aligned} P (4, 4) \end{aligned}$
::P(4 4 4 4 )

$\begin{aligned} P (5, 3) \end{aligned}$
:P 5 , 3 )

$\begin{aligned} P (8, 4) \end{aligned}$
:P 8 , 4 )

$\begin{aligned} P (9, 4) \end{aligned}$
::P(9,4)

$\begin{aligned} P (10, 3) \end{aligned}$
::P(10 , 3 )

$\begin{aligned} P (12, 2) \end{aligned}$
::P(12 12 , 2 )

$\begin{aligned} P (9, 3) \end{aligned}$
::P(9,3)

$\begin{aligned} P (8, 6) \end{aligned}$
:P 8 , 6 )

$\begin{aligned} P (9, 3) \end{aligned}$
::P(9,3)

$\begin{aligned} P (10, 3) \end{aligned}$
::P(10 , 3 )

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。

Section outline

General

排列

Finding Permutations ::查找差异

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