课程： 13.5 合并 | The Way To Learn

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合并

Combinations
::合并

In the previous lesson we looked at situations where the order of an arrangement is important. For example, when looking at 3-digit numbers, the number 123 is very different from the number 312, even though it contains the same 3 digits. But in some situations, the order is not important – for example, when looking at cards in a hand of poker, or choosing toppings to put on a pizza. In these situations, when order is not important , we are looking at combinations of items. For example, if you are a poker player you might want to know the probability of being dealt four aces in a hand of five cards. You don’t care in which order you receive the ace cards – only the fact that you have four of them is important.
::在上一课中,我们审视了安排顺序很重要的情况。比如,当看3位数数字时,123号与312号数字大不相同,尽管3位数相同。但在某些情况下,顺序并不重要 — — 比如,当用扑克手看牌牌时,或者选择把图纸放在比萨饼上时。在这些情况下,当顺序不重要时,我们审视的是项目组合。比如,如果你是扑克玩家,你也许想知道用5张牌拿4张A的可能性。你并不关心获得A卡的顺序 — — 只有4张才是重要的事实。

Counting Combinations
::计算组合

Just as with permutations, it’s sometimes easiest to calculate combinations by listing all the possibilities available, and counting them. The only difference is, when we list one , we automatically exclude a larger number of permutations. For example a poker hand that is (ace, ace, ace, ace, king $\begin{aligned} ♣ \end{aligned}$ ) is identical to (ace, ace, ace, king $\begin{aligned} ♣ \end{aligned}$ , ace), (ace, ace, king $\begin{aligned} ♣ \end{aligned}$ , ace, ace), (ace, king $\begin{aligned} ♣ \end{aligned}$ , ace, ace, ace) and (king $\begin{aligned} ♣ \end{aligned}$ , ace, ace, ace, ace). So we must be careful to use a listing method that includes all combinations without repeating ones that are really the same. Let’s examine a situation where it’s relatively straightforward to do that.
::和变相一样,通过列出所有的可能性来计算组合有时也比较容易。唯一的区别是,当我们列出一个时,我们自动排除了更多的变相。比如,牌手(Ace、Ace、Ace、Ace、Ace、Ken)与(Ace、Ace、Ken、Ace、Ace、Ace、Ace、Ace、Ace、Ace、Ace、Ace、Ace、Ace、Ace、Ace、Ace、Ace、Ace、Ace等)完全相同。因此,我们必须小心地使用包括所有组合的列表方法,而不必重复那些完全相同的组合。让我们来研究一个相对简单的情况。

Anne wishes to knit herself a striped sweater. She has 4 colors of yarn available; red, blue, green and yellow. How many different combinations of two colors does she have to choose from?
::安妮想给自己织件条纹毛衣。她有4种线条:红色、蓝色、绿色和黄色。她需要选择多少种两种颜色的不同组合?

When we just choose color pairs, there will be fewer combinations than we would have if we were counting permutations as in the previous lesson. For example red and blue is equivalent to blue and red , and we should only count one as a unique pairing. We start by listing the color pairs but we will also write down equivalent pairings at the same time. This will help prevent us from repeating combinations:
::当我们只选择颜色配对时, 组合会比之前的课数的组合少。例如, 红色和蓝色相当于蓝色和红色, 我们只应该将一个配对计算为独特的配对。我们从列出颜色配对开始, 但是我们也会同时写下等配对。这将帮助我们避免重复组合 :

Pairing	Equivalent pairings (do not count)
Red & blue	blue & red
Red & green	green & red
Red & yellow	yellow & red
Green & blue	blue & green
Green & yellow	yellow & green
Yellow & blue	blue & yellow

So there are 6 distinct combinations. There are also 6 “repeat” pairings – for every pair of colors we choose there is 1 combination but 2 permutations. Anne can choose from six distinct color pairs for her sweater .
::所以有6种不同的组合。还有6种“重复”配对 — — 我们选择的每对颜色都有1种组合,但有2种变异。安妮可以从6种不同的颜色中选择她的毛衣。

Real-World Application: Pizza Toppings
::真实世界应用程序:披萨托普

Triominoes Pizza Company specializes in 3-topping pizzas. If the available toppings are cheese, pepperoni, mushroom, pineapple and olives, how many different 3-topping combinations can customers choose from?
::Triominoes Pizza公司专门从事3吨比萨饼。如果可用药头是奶酪、辣椒、蘑菇、菠萝和橄榄,顾客可以选择多少种3吨组合?

We’ll start by making a table and list first choice, second choice and third choice:
::首先, 我们先做一张表格, 列出第一选择、第二选择和第三选择:

$\begin{aligned} 1^{s t} \end{aligned}$ topping	$\begin{aligned} 2^{n d} \end{aligned}$ topping	$\begin{aligned} 3^{r d} \end{aligned}$ topping
cheese	pepperoni	mushroom
cheese	pepperoni	pineapple
cheese	pepperoni	olives
cheese	mushroom	pineapple
cheese	mushroom	olives
cheese	pineapple	olives
pepperoni	mushroom	pineapple
pepperoni	mushroom	olives
pepperoni	pineapple	olives
mushroom	pineapple	olives

Note that as we progress through the choices for first topping, the number of combinations we have for the second and third toppings get fewer. This is because some combinations have already been used, in a different order.
::请注意,随着我们通过选择第一次加注,我们第二和第三次加注的组合数量越来越少。这是因为一些组合已经被不同顺序地使用过。

By counting the table entries we see that there are 10 possibilities for a 3-topping pizza.
::通过计算表格条目,我们看到有10种可能进行3吨比萨饼。

Determining Combinations by Looking at Permutations
::通过审视差异来确定组合

You can see that there are always fewer combinations than permutations in a given situation – but you should also see that knowing which combinations have already been used is important to avoid counting combinations twice. One combination can give rise to several permutations of the same objects.
::你可以看到,在特定情况下,组合总是比变异少 — — 但你也应该看到,知道哪些组合已经使用过,对于避免两次计算组合很重要。一种组合可能导致相同对象的若干变异。

Another way to calculate combinations is this: If we know how many permutations there are in a system and how many permutations there are for each combination , then we can divide the number of permutations by the number of permutations for each combination to get the number of combinations .
::计算组合的另一种方法是:如果我们知道一个系统中有多少变异,每个组合有多少变异,那么我们可以将变异次数除以每个组合的变异次数,以获得组合的数目。

To illustrate this, look again at the Triominoes Pizza menu. If we were to look at permutations of toppings, we could quickly calculate that there are $\begin{aligned} 5 \times 4 \times 3 = 60 \end{aligned}$ permutations. But for every 3-topping choice there are several permutations which are all equivalent. For example, the following all give identical pizzas:
::为了说明这一点,请再看看三一式披萨菜单。如果我们查看图案的排列,我们可以快速计算出5x4x3=60的排列。但是,对于每3个位数的选择,就有一些等同的排列。例如,以下各点都给出相同的披萨:

\begin{aligned} Cheese, pepperoni & mushroom Cheese, mushroom & pepperoni \\ Pepperoni, cheese & mushroom Pepperoni, mushroom & cheese \\ Mushroom, cheese & pepperoni Mushroom, pepperoni & cheese \end{aligned}

::奶酪、辣椒尼和蘑菇芝士、蘑菇和辣椒皮帕罗尼、奶酪和蘑菇辣椒尼、蘑菇和芝士蘑菇、奶酪蘑菇和芝士蘑菇、奶酪和辣椒尼蘑菇、辣椒和奶酪

So for every different combination there are 6 distinct permutations. Since we have a formula for counting permutations, we can use it to find out how many total permutations there are and simply divide that number by 6:
::因此,对于每一个不同的组合,都有6个不同的组合。由于我们有一个计算排列的公式,我们可以用它来找出有多少个总排列,而仅仅将这个数字除以6:

\begin{aligned} Combinations = \frac{1}{6} \cdot_{5} P_{3} = \frac{1}{6} \cdot \frac{5!}{2!} = \frac{1}{6} \cdot \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = \frac{60}{6} = 10 \end{aligned}

::组合=165P3=165!5! 2!=165x4x4x3x2x12x1=606=10

Find Combinations Using a Formula
::使用公式查找组合

If you look back at examples 1 and 2 you can see that the number of permutations is a simple multiple of the number of combinations. In example 1 there are two times as many permutations as combinations. In example 2 there are six times as many permutations as combinations. One question you might be asking is: “How do I know how the number of combinations is related to the number of permutations?”
::回顾例1和例2,你可以看到变相的数量是组合数的简单倍数。在例1中,变相数是组合数的两倍。在例2中,变异数是组合数的六倍。你可能要问的一个问题是:“我怎么知道组合数与变异数数有何关系?”

A more important question might be “where did the numbers two and six come from?” If you think carefully you should realize that any time you’ve chosen 2 objects, there are only 2 ways of ordering them, while if you’ve chosen 3 objects there are 3! ways of ordering them (6 ways). Similarly, if you choose 7 objects, there will be 7! ways of arranging them. We can make use of this fact when calculating combinations. The number of combinations is the number of permutations divided by the number of ways of arranging the items you choose. If you choose $\begin{aligned} r \end{aligned}$ objects from a collection of $\begin{aligned} n \end{aligned}$ objects there are $\begin{aligned} r! \end{aligned}$ ways of arranging what you choose. In equation form, the number of combinations is therefore:
::一个更重要的问题可能是“数字二和数字六来自何处?” 如果你仔细思考,你应该意识到,你选择了2个对象,那么只要选择了2个对象,就只有2种订购方法,而如果选择了3个对象,就有3个!命令它们的方法(6种方法)。同样,如果选择了7个对象,就有7个方法!安排它们的方法。在计算组合时,我们可以利用这个事实。组合的数目是调和次数,除以选择项目安排方式的数量。如果从n对象的集合中选择 R 对象,则有r 方法来安排你选择的东西。在等式形式中,组合的数目是:

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

::来! 来啊! 来啊! 来啊! 来啊! 来啊! 来啊! 来啊!

In other words, the number of combinations is equal to the number of permutations divided by $\begin{aligned} r! \end{aligned}$ (because $\begin{aligned} r! \end{aligned}$ is the number of permutations for each combination). We can use this new formula to quickly calculate combinations without listing them all.
::换句话说,组合的数量等于除以 r!( 因为 r. 是每个组合的变异数量 ) 。我们可以使用这个新公式快速计算组合, 而不列出所有组合。

Andrew is packing for a vacation. He owns twelve shirts and wants to pack five of them. How many combinations of shirts does he have to choose from?
::他有十二件衬衫,想打五件

Since the order he packs his shirts in is not important, we are looking at a combination. He is choosing five shirts from a total of twelve:
::因为他把衬衫打包的顺序并不重要,我们正在看一个组合。他从总共12件中选择了5件衬衫:

Choosing 5 from 12: $\begin{aligned} _{12} C_{5} = \frac{12!}{(12 - 5)! 5!} = \frac{12!}{7! 5!} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792 combinations. \end{aligned}$
::从 12: 12C5=12! (12- 5)! 5! 12! 7! 5! 12! 7! 5! = 12x11x10x9x85x4x3x21=792 组合中选择 5 。

Example
::示例示例示例示例

Example 1
::例1

At Summerfield High School, the student council has 8 students, 3 of whom are needed to be on the prom committee. How many ways can the prom committee be chosen?
::在夏季菲尔德高中,学生委员会有8名学生,其中3人需要参加舞会委员会。

There are:
::它们是:

\begin{aligned} _{n} C_{r} & = \frac{n!}{(n - r)! r!} \\ _{8} C_{3} & = \frac{8!}{(8 - 3)! 3!} \\ _{8} C_{3} & = \frac{8!}{5! 3!} \\ _{8} C_{3} & = \frac{8 \cdot 7 \cdot 6}{3!} \\ _{8} C_{3} & = 8 \cdot 7 = 56 \end{aligned}

::nCr=n!r! 8C3=8! (8- 3)!3!8C3=8!5!3!3!8C3=8C3=8!5!3!8C3=8(7)63!8C3=8_7=56

56 ways to chosen the prom committee.
::56种选择舞会委员会的方法

Review
::回顾

For 1-3, how many combinations are possible in the following situations?
::对于1-3,在以下情况下有多少组合是可能的?

Buying a hot-dog with two of the following toppings: ketchup, mustard, chili, cheese, & pickles.
::购买热狗,配有以下两种托普:番茄酱、芥末、辣椒、奶酪和泡菜。

Choosing 5 CDs from a selection of 8.
::从所选的8张中选择5张CD。

Selecting 3 games from a box containing Scrabble, Twister, Connect-4, Snap & Mousetrap.
::从含有拼字游戏的盒子中选择 3 个游戏, Twister, 连接 4, 连接 4, Snap & 鼠标陷阱。

$\begin{aligned} _{3} C_{1} \end{aligned}$
::3C1 3C1

$\begin{aligned} _{7} C_{1} \end{aligned}$
::7C1 7C1

$\begin{aligned} _{6} C_{2} \end{aligned}$
::6C2 6C2

$\begin{aligned} _{8} C_{8} \end{aligned}$
::8C8

$\begin{aligned} _{9} C_{3} \end{aligned}$
::9C3(9C3)

$\begin{aligned} _{9} C_{6} \end{aligned}$
::9C6

$\begin{aligned} _{7} C_{3} \end{aligned}$
::7C3 7C3

$\begin{aligned} _{17} C_{4} \end{aligned}$
::17C4 17C4

$\begin{aligned} _{30} C_{11} \end{aligned}$
::30C11 30C11

$\begin{aligned} _{10} C_{0} \end{aligned}$
::10C0 10C0

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

合并

Combinations ::合并

Counting Combinations ::计算组合

Real-World Application: Pizza Toppings ::真实世界应用程序:披萨托普

Determining Combinations by Looking at Permutations ::通过审视差异来确定组合

Find Combinations Using a Formula ::使用公式查找组合

Example ::示例示例示例示例

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)