课程： 12.5 定义和应用组合

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定义和应用组合

A lottery system consists of 40 balls numbered 1 through 40. In the Big 5, five balls are selected from these 40 balls. You need to match all five numbers to win the cash prize. How many ways can 5 balls from a selection of 40 numbers be drawn?
::彩票系统由40个球组成,编号为1至40。在大五中,从这40个球中挑选出5个球。你需要匹配所有5个数字才能赢得现金奖。从选择的40个数字中抽出5个球可以抽出多少方法?

Combinations
::合并

Combinations of a subset of a larger set of objects refer to the number of ways we can choose items in any order. For comparison, look at the table below to see when order matters and when order doesn’t matter.
::较大一组天体的子集组合是指我们可以按任何顺序选择项目的方式的数量。比较时,请看下表,看顺序问题和顺序无关紧要的时间。

Combinations	Permutations
Ways to select the members of a committee from a larger population ::从较大人口中选出委员会成员的方法	Ways to select specific officers in a club-president, vice president, treasurer, etc. ::如何挑选俱乐部主席、副主席、司库等具体官员。
Ways to select a set number of pizza toppings from a larger list of choices ::从更大的选择列表中选择一组比萨饼比萨饼比点数的方法	Ways to select and arrange scoops of ice cream on a cone ::选择和安排冰淇淋冰淇淋在锥锥上的方法
Ways to select books from a reading list ::从阅读列表中选择书籍的方法	Ways to select and order in which to read books selected from a reading list. ::选择和顺序阅读从阅读列表中选择的书籍的方法。

The simplest way to describe the difference between a and a is to say that in a combination the order doesn’t matter. The members of a committee could be selected in any order but the officers in a club are assigned a specific position and therefore the order does matter. Be careful of the use of these words in the real world as they are sometimes misused. For example, a locker combination. The ways to select and order the three numbers for a locker combination is not actually a combination, but a permutation since the order does matter.
::描述a和a之间的区别的最简单方式就是说,在组合中,顺序并不重要。委员会成员可以按任何顺序选出,但俱乐部的军官可以被指定一个特定的职位,因此命令很重要。当这些词在现实世界中被使用时, 当它们有时被滥用时要小心使用。例如, 储物柜组合。选择和订购储物柜组合的三个数字的方法实际上不是一个组合, 而是一个调和, 因为命令很重要。

Let's solve the following problem using combinations.
::让我们用组合来解决以下的问题。

How many ways can we choose three different flavors of ice cream from a selection of 15 flavors to place in a bowl?
::我们从15种口味中选择三种不同口味的冰淇淋来投放在碗里?

First, does order matter in the bowl? When we created an ice cream cone with three scoops in an earlier concept the order did matter but here it does not. Let’s work from the example of the ice cream cone. We determined the number of permutations of a subset of three flavors from the total 15 flavors using the formula: $\begin{align*}\frac{n!}{(n-r)!}=\frac{15!}{(15-3)!}=\frac{15!}{12!}=\frac{15\times14\times13\times{\color{red}\cancel{12!}}}{{\color{red}\cancel{12!}}}=2730\end{align*}$ . Now that order doesn’t matter, this number includes the $\begin{align*}3!\end{align*}$ ways to arrange each combination of 3 flavors. We can divide 2,730 by $\begin{align*}3!\end{align*}$ to determine the number of combinations: $\begin{align*}\frac{2730}{3\times2\times1}=\frac{2730}{6}=455\end{align*}$ .
::首先, 顺序是否在碗中重要 ? 当我们用前一个概念中的三个勺子创建了一个冰淇淋锥时, 顺序并不重要, 但在这里它并不重要。让我们从冰淇淋锥的例子中来研究一下。我们用公式确定了15种口味中三种口味的分数: n! ! 15!! (15- 3) ! (15- 3) ! =15! 12! =15. 12! =15x14x13x13x12! 12! =2730。现在这个顺序并不重要, 这个数目包括3种口味组合的3种方式。我们可以将2, 730 乘 3! 来决定组合数 : 27303x2x1= 27306= 455 。

The notation and formula for combinations can be written as: $\begin{align*}\dbinom{n}{r}={_nC}_r=C^n_r=\frac{n!}{r!(n-r)!}\end{align*}$ , where $\begin{align*}n\end{align*}$ represents the number of elements in the set and $\begin{align*}r\end{align*}$ represents the number of elements in the subset.
::组合的符号和公式可以写为nr) =nCr=Cnr=n!r!(n-r!) , n 表示集中元素的数量, r 表示子集中元素的数量。

Now, let's evaluate the following expressions.
::现在,让我们来评估以下表达式。

$\begin{align*}\dbinom{8}{5}\end{align*}$

$\begin{align*}\dbinom{8}{5}=\frac{8!}{5!(8-5)!}=\frac{8\times7\times{\color{red}6}\times{\color{red}\cancel{5!}}}{{\color{red}\cancel{5!}}\times{\color{red}3}\times{\color{red}2}\times1}=\frac{8\times7}{1}=56\end{align*}$ .

$\begin{align*}_8C_0\end{align*}$
::8C0 8C0

Type in 8 on the TI-83 Graphing calculator, then MATH $\begin{align*}\rightarrow\end{align*}$ PRB, select 3: $\begin{align*}_nC_r\end{align*}$ . Now type in 0 and your screen should read 8 $\begin{align*}_nC_r\end{align*}$ 0 before your press ENTER to get the answer 1.
::在 TI-83 图形计算器上输入 8 的 8, 然后 MATH PRB, 选择 3 : nCr. 现在输入 0 。您的屏幕应该在按 ENTER 获得答案 1 之前读为 8 nCr 0 。

$\begin{align*} _8C_8\end{align*}$
::8C8

Type in 8 on the TI-83 Graphing calculator, then MATH $\begin{align*}\rightarrow\end{align*}$ PRB, select 3: $\begin{align*}_nC_r\end{align*}$ . Now type in 8 and your screen should read 8 $\begin{align*}_nC_r\end{align*}$ 8 before your press ENTER to get the answer 1.
::在 TI-83 绘图计算器上输入 8, 然后 MATHPRB, 选择 3 : nCr。现在输入 8 。您的屏幕应该读8 nCr 8, 然后再按 ENTER 获得答案 1 。

$\begin{align*}C^{10}_7\end{align*}$
::C107 C107

$\begin{align*}C^{10}_7=\frac{10!}{7!(10-7)!}=\frac{10\times9\times8\times{\color{red}\cancel{7!}}}{{\color{red}\cancel{7!}}\times3!}=\frac{10\times{\color{red}\cancel{3}}\times3\times4\times{\color{red}\cancel{2}}}{{\color{red}\cancel{3}}\times{\color{red}\cancel{2}}}=120\end{align*}$
::C107=1010! 7! (10- 7) =10x9x8x7! 7x7! 7x3! =10x3x3x3x4x23x2=120

Explain why the answers to #2 and #3 are the same.
::解释为何二号与三号的答案相同。

In problem 2 we are looking at the ways to choose 0 items from 8 choices. There is only one way to do this. In problem 3 we are looking at the ways to choose 8 items from 8 choices. Well, the only way to do that is to choose all 8 items. So, there is only 1 way to choose zero items or all the items from a set.
::在问题2中,我们正在研究从8个选项中选择0个项目的方法。只有一种方法可以做到这一点。在问题3中,我们正在研究从8个选项中选择8个项目的方法。唯一的方法就是选择所有8个项目。因此,只有一种方法可以从一组中选择零项目或所有项目。

All of the notations in problems 1-4 indicate that we should use the formula for a combination. We can use the graphing calculator to evaluate these as well. Problems 2 and 3 are set up in the form of the calculator notation so we will use the calculator to evaluate those two and the formula for the other two.
::问题1-4中的所有标记都表示我们应该使用公式来组合。我们可以使用图形计算器来评估这些。问题2和问题3以计算符号的形式设置, 这样我们就可以使用计算器来评估这两个公式和另外两个公式。

Finally, let's solve the following problem using combinations.
::最后,让我们用组合来解决以下的问题。

How many ways can a team of five players be selected from a class of 20 students?
::从20个学生班中选出5个球员,

We can express this problem using the notation $\begin{align*}\dbinom{20}{5}\end{align*}$ and then use the formula to evaluate.
::我们可以用标记(205)来表示这个问题,然后用公式来评价。

$\begin{align*}\dbinom{20}{5}=\frac{20!}{5!\times15!}=\frac{{\color{red}\cancel{20}}\times19\times{\color{red}\cancel{6}}\times3\times17\times16\times{\color{red}\cancel{15!}}}{{\color{red}\cancel{5\times4}}\times{\color{red}\cancel{3\times2}}\times{\color{red}\cancel{15!}}}=15,504.\end{align*}$

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the number of ways 5 balls can be drawn from a selection of 40 numbers.
::早些时候,有人要求你从选择的40个数字中找到5个球可以抽出的方法的数目。

First, we need to determine whether order matters. The winning numbers 5, 10, 15, 20, and 25 are the same as the winning numbers 25, 5, 20, 10, 15, so order does not matter.
::首先,我们需要确定顺序是否重要。胜者人数是5, 10, 15, 20, 25, 和胜者人数是25, 5, 20, 10, 15, 15, 因此,胜者人数是25, 20, 10, 10, 15, 5。因此,胜者人数与胜者人数是相同的。

Therefore, we use the combinations formula $\begin{align*}\dbinom{n}{r}=\frac{n!}{r!(n-r)!}\end{align*}$
::因此,我们使用组合公式(nr)=n!r!r!!

$\begin{align*}\frac{40!}{5!(40-5)!}\\ \frac{40!}{5!35!}\\ \frac {40\cdot39\cdot38\cdot37\cdot36\cdot35!}{5\cdot4\cdot3\cdot2\cdot1\cdot 35!}\\ \frac{78,960,960}{120} = 258,008\end{align*}$

Therefore, there are 258,008 ways five winning balls can be drawn from the 40 numbers.
::因此,从40个数字中可以抽出5个赢球的258 008种方法。

For Examples 2-4, evaluate using the formula for combinations of the calculator.
::关于实例2-4,使用计算器组合的公式进行评价。

Example 2
::例2

$\begin{align*}\dbinom{7}{5}\end{align*}$

$\begin{align*}7 \ _nC_r \ 5 = 21\end{align*}$
::7 nCr 5=21

Example 3
::例3

$\begin{align*}_{20}C_{12}\end{align*}$
::20C12 20C12

$\begin{align*}20 \ _nC_r \ 12 = 125,970\end{align*}$
::20 nCr 12=125 970

Example 4
::例4

$\begin{align*}C^{15}_7\end{align*}$
::C157 C157

$\begin{align*}15 \ _nC_r \ 7 = 6,435\end{align*}$
::15 nCr 7=6 435

Example 5
::例5

How many ways can a committee of three students be formed from a club of fifteen members?
::从一个由十五名成员组成的俱乐部中组成由三名学生组成的委员会,可以采用多少方式?

$\begin{align*}\dbinom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{{\color{red}\cancel{3}}\times5\times{\color{red}\cancel{2}}\times7\times13\times{\color{red}\cancel{12!}}}{{\color{red}\cancel{3\times2}}\times{\color{red}\cancel{12!}}}=455\end{align*}$ .

Example 6
::例6

How many three-topping pizzas can be made if there are 10 topping choices?
::如果有10个3吨的比萨饼选择的话,可以做多少个3吨的比萨饼?

$\begin{align*}C^{10}_3=10 \ _nC_r \ 3 = 120\end{align*}$ .
::C103=10 nCr 3=120。

Review
::回顾

Evaluate the following combinations with or without a calculator.
::评估以下与计算器或不与计算器的组合。

$\begin{align*}_{13}C_{10}\end{align*}$
::13C10 13C10

$\begin{align*}C^{10}_6\end{align*}$
::C106 C106
$\begin{align*}\dbinom{18}{10}\end{align*}$

Explain why $\begin{align*}_9C_5={_9C}_4=126\end{align*}$ .
::解释为什么9C5=9C4=126。

Decide whether the following situations are permutations and which are combinations.
1. Ways to arrange students in a row.
  ::连续安排学生的方式。
2. Ways to select a group of students.
  ::选择一组学生的方法。
3. Ways to organize books on a shelf.
  ::如何在架子上组织书籍。
4. Ways to select books to read from a larger collection.
  ::选择从较大收藏中读取书籍的方法。
5. Ways to select three different yogurt flavors from a collection of ten flavors.
  ::选择三种不同的酸奶口味的方法从十种口味的集合中挑选出来
::决定下列情况是否是变式和组合。排列学生顺序的方法; 选择一组学生的方法; 在架子上组织书籍的方法; 从更大的收藏中选择书籍来阅读的方法。从10种口味的集合中选择三种不同的酸奶口味的方法。

In each scenario described below, use either a combination or permutation as appropriate to answer the question.
::在下文所述每一种情况中,酌情使用组合或变式来回答问题。

There are seven selections for appetizers on a caterer’s menu. How many ways can you select three of them?
::餐饮菜单上有七种开胃菜选择。您可以选择其中三种方式吗 ?

You only have time for seven songs on your workout playlist. If you have 10 favorites, how many ways can you select seven of them for the list? Now, how many ways can you select them in a particular order?
::您在游戏列表中只有7首歌的时间。如果您有 10 个最喜爱的曲目, 您可以为列表选择 7 个曲目。您可以选择多少种方式来选择 7 个曲目 ? 现在, 您可以选择多少种方式来选择这些曲目 ?

How many ways can you select two teams of five players each from a group of ten players?
::你从十个球员的一组中选择由五个球员组成的两个球队有多少种方法?

At the local frozen yogurt shop a sundae comes with your choice of three toppings. If there are 12 choices for toppings, how many combinations of toppings are possible?
::在当地的冷冻酸奶店,圣代会带来你选择的3吨。如果有12种选择,那么可以使用多少种组合?

How many ways can four people be selected from a group of 30 to serve on a committee? What if each of the four people was selected to fill a specific position on the committee?
::从30人中挑选4人担任委员会成员,可以选用多少方法?如果选用4人中的每一人填补委员会的一个具体职位呢?

A soccer team has 20 players, but only 11 play at any one time.
1. How many ways can the coach select a group of eleven players to start (disregard positions)?
  ::教练可以选择多少种方式来选择一组由11名球员组成的球员来开始(忽略位置)?
2. Now, of the eleven players on the field, one is a goalie, four play defense, three play midfield and three play offense. How many ways are there to assign the eleven players to these positions?
  ::现在,在球场的11个球员中,一个是守门员,4个是防守,3个是中场球员,3个是进攻。
3. Considering your answers to parts a and b, how many ways can the coach select eleven players and assign them positions on the field? Assume all players can play each position.
  ::考虑到你对A和B部分的回答, 教练可以选择多少种方法来选择11个球员, 并分配他们到球场上的位置? 假设所有球员都可以扮演每个位置。
::足球队有20个球员, 但每次只有11个球员。教练可以选择多少种方法来选择一组11个球员来开始( 忽略位置 ) ? 现在, 在球场的11个球员中, 1个是守门员, 4个是守球员, 3个是中场员, 3个是球员。有多少方法可以分配11个球员到这些位置? 考虑到你对A和b部分的回答, 教练可以选择多少种方法来选择11个球员, 并在球场上分配他们的位置? 假设所有球员都可以在每个球员的位置上打球。

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

定义和应用组合

Combinations ::合并

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Review ::回顾

Review (Answers) ::回顾(答复)