Course: 8.6 计算合并二 | The Way To Learn

The Way To Learn

Section outline

Select section General

Collapse Expand
General

Collapse all Expand all
- Select activity 新闻通告
  
  新闻通告 Forum
Select section 计算合并二

Collapse Expand
计算合并二
How many different ways could you select three coins from a pile of coins containing quarters, dimes, nickels, and pennies, if you can take as many of each kind as you like to make up the total of three?
::从一堆硬币中挑选三个硬币有多少种不同的方法? 硬币,包括硬币、硬币、硬币、硬币、硬币和硬币,如果你能拿多少种来组成总共三个?

At the end of the lesson, we’ll return to this problem and see if we can figure it out.
::学习结束时, 我们会回到这个问题上来, 看看我们能否找出答案。

Calculating Combinations: Special Cases
::计算合并:特殊情况

Counting combinations when some members of the source set either cannot be told apart or may be used multiple times is a bit different from calculating them otherwise. In order to practice these types of calculations, we are going to use combination notation , which is similar to the permutation notation that we discussed in a prior lesson.
::当源集的一些成员要么无法分解,要么可能被多次使用时,计算组合与其他计算略有不同。为了实践这些类型的计算方法,我们将使用组合符号,这类似于我们在前一课中讨论过的变换符号。

There are two common types of combination notation : $\begin{aligned} _{n} C_{r} \end{aligned}$ and $\begin{aligned} C (n, r) \end{aligned}$
::有两种常见的混合编号:nCr和C(n,r)

$\begin{aligned} n = \end{aligned}$ number of different units to choose from
::n=选择的不同单位数

$\begin{aligned} r = \end{aligned}$ number of units in each group.
::r=每个组的单位数目。

Once you know $\begin{aligned} n \end{aligned}$ and $\begin{aligned} r \end{aligned}$ , you can use the formula we reviewed in Calculating Combinations I if your problem involves combinations without repeats :
::一旦您知道 n和 r, 您可以使用我们在计算组合中审查的公式I, 如果您的问题涉及组合, 而不重复 :

$\begin{aligned} \frac{n!}{(n - r)! \times r!} \end{aligned}$

::不! ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

However, to calculate combinations with repetition allowed, you need a different formula:
::然而,要计算重复允许的组合,您需要不同的公式:

$\begin{aligned} \frac{(n + r - 1)!}{(n - 1)! \times r!} \end{aligned}$

:n+r-1)!(n-1)!xr!

Scooping Ice Cream
::冰淇淋

How many unique three-scoop bowls of ice cream are possible with twelve flavors to choose from?
::有多少独一无二的三碗冰淇淋可以选择十二口味的冰淇淋?

This is a combination question, with repeated choices allowed (it is OK to have a bowl with three scoops of chocolate). Since there are 12 choices, and we choose 3, the notation is: $\begin{aligned} _{12} C_{3} \end{aligned}$ , and we need to use the formula for repeat-allowed combinations: $\begin{aligned} \frac{(n + r - 1)!}{(n - 1)! \times r!} \end{aligned}$ . Here, $\begin{aligned} n = 12 \end{aligned}$ and $\begin{aligned} r = 3 \end{aligned}$ .
::这是一个组合问题, 重复选择是允许的( 有三勺巧克力的碗是允许的 ) 。由于有12个选择, 我们选择了 3 , 符号是 12C3 , 我们需要使用重复允许组合的公式 : (n+r-1) ! (n-1)! xr! 这里, n=12 和 r=3 。这里, n=12 和 r= 3 。

$\begin{aligned} \frac{(12 + (3 - 1))!}{(12 - 1)! \times 3!} \\ \frac{14!}{11! \times 3!} \\ \frac{14 \times 13 \times 12}{3 \times 2 \times 1} \\ \frac{2184}{6} \\ 364 u n i q u e b o w l s o f i c e c r e a m \end{aligned}$

:12+(3- 1) ! (12+( 3- 1) !x3! 14. 11! x3! 14x13x123xx2x121846) 364 独特冰淇淋碗

Creating Teams
::创建团队

How many unique teams of four can be made from thirteen people?
::13个人能组成多少个独特的四人小组?

This is a combination without repetition question (the same person cannot be two different members of the team). Since there are thirteen different people, and we are choosing four, the notation is $\begin{aligned} _{13} C_{4} \end{aligned}$ , and we need to use the combination without repetition formula: $\begin{aligned} \frac{n!}{(n - r)! \times r!} \end{aligned}$ , where $\begin{aligned} n = 13 \end{aligned}$ and $\begin{aligned} r = 4 \end{aligned}$ .
::这是一个不重复问题的组合( 同一人不能是团队中两个不同的成员 ) 。由于有13个不同的人, 我们选择了4个, 标记是13C4, 我们需要使用组合而不重复公式 : n! (n-r)! xr!, 其中 n=13 和 r=4 。

$\begin{aligned} \frac{13!}{(13 - 4)! \times 4!} \\ \frac{13!}{9! \times 4!} \\ \frac{13 \times 12 \times 11 \times 10}{4!} \\ \frac{17160}{24} \\ 715 t e a m s \end{aligned}$

::13号 13号 14号 13号 13号 9号 9号 13号 13号 12号 11号 104号 17号 16024715小组

Handfuls of Candy
::甜甜甜甜甜甜甜甜

How many unique handfuls of candy, containing less than five pieces each, are possible with nine flavors to choose from?
::有多少独一无二的糖果,每颗糖的含量不到五块, 可以用九种口味来选择?

Since the problem specifies “less than five pieces”, we need to complete multiple steps because we need to know how many combinations (repetition allowed) are possible from four, three, two, and one piece of candy. For each number, we need to calculate the number of possible combinations, using the “repetition allowed” formula: $\begin{aligned} \frac{(n + r - 1)!}{(n - 1)! \times r!} \end{aligned}$
::由于问题具体说明了“小于五块”的问题,我们需要完成多个步骤,因为我们需要知道从四、三、二和一块糖果中有多少组合(允许重复)是可能的。对于每个数字,我们需要使用“允许重复”公式计算可能的组合数量n+r-1)!(n-1)!xr!

One piece: This is easy, one possible option from each of nine flavors: 9 combinations
::一件:这很容易,一个可能的选项来自9种口味:9种组合

Two pieces: $\begin{aligned} _{9} C_{2} = \frac{(9 + 2 - 1)!}{(9 - 1)! \times 2!} = \frac{10!}{8! \times 2!} = \frac{10 \times 9}{2 \times 1} = \frac{90}{2} = 45 c o m b i n a t i o n s \end{aligned}$
::两块: 9C2=( 9+2- 1) ! (9- 1) ! x2! = 10! 8! x2! = 10x92x1=902=45 组合

Three pieces: $\begin{aligned} _{9} C_{3} = \frac{(9 + 3 - 1)!}{(9 - 1)! \times 3!} = \frac{11!}{8! \times 3!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = \frac{990}{6} = 165 c o m b i n a t i o n s \end{aligned}$
::三块: 9C3=( 9+3- 1) ! (9- 1) ! x3! = 11! 8! x3! = 11x10x93xxx2x1=9906=165 组合

Four pieces: $\begin{aligned} _{9} C_{4} = \frac{(9 + 4 - 1)!}{(9 - 1)! \times 4!} = \frac{12!}{8! \times 4!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \frac{11880}{24} = 495 c o m b i n a t i o n s \end{aligned}$
::四块: 9C4=9C4=( 9+4- 1) ! (9- 1) ! x4! =12! 8! x4! =12x11x10x10x94x3x3x2x1=1188024=495 组合

$\begin{aligned} TOTAL = 9 + 45 + 165 + 495 = 714 p o s s i b l e h a n d f u l s \end{aligned}$

::共计=9+45+165+495=714

Earlier Problem Revisited
::重审先前的问题

How many different ways could you select three coins from a pile of coins containing quarters, dimes, nickels, and pennies, if you can take as many of each kind as you like to make up the total of three?
::从一堆硬币中挑选三个硬币有多少种不同的方法? 硬币,包括硬币、硬币、硬币、硬币、硬币和硬币,如果你能拿多少种来组成总共三个?

Now that we have completed a few similar examples, you should recognize that this could be written as a $\begin{aligned} _{4} C_{3} \end{aligned}$ question, with repeats allowed, meaning that we should use the formula: $\begin{aligned} \frac{(n + r - 1)!}{(n - 1)! \times r!} \end{aligned}$
::现在我们已经完成了几个类似的例子,你应该认识到,这可以写成4C3问题,但允许重复,这意味着我们应该使用公式n+r-1)!(n-1)!xr!

$\begin{aligned} n = 4 \end{aligned}$ and $\begin{aligned} r = 3 \end{aligned}$
::n=4和r=3

$\begin{aligned} \frac{6!}{3! \times 3!} \\ \frac{6 \times 5 \times 4}{6} \\ \frac{120}{6} \\ 20 w a y s \end{aligned}$

::6! 3! x3! 6x5x46120620 方式

Examples
::实例

How many ways can a five-person team be chosen from twenty-three people?
::从23人中选出一个五人小组可选用多少方法?

How many three-color combinations can be selected from the seven-color basic rainbow?
::从七色基本彩虹中可以选取多少三色组合?

How should $\begin{aligned} _{8} C_{2} \end{aligned}$ be read? How many combinations does it indicate?
::如何读8C2?它表示有多少组合?

Solutions :
::解决办法:

Example 1
::例1

How many ways can a five-person team be chosen from twenty-three people?
::从23人中选出一个五人小组可选用多少方法?

There are 23 people to choose from, so $\begin{aligned} n = 23 \end{aligned}$ . We are choosing members in groups of 5, so $\begin{aligned} r = 5 \end{aligned}$ . Use the formula for non-repeating combinations: $\begin{aligned} \frac{n!}{(n - r)! \times r!} \end{aligned}$
::共有23个人需要选择, 所以 n=23。我们选择的是5个组的成员, 所以 r=5 。使用非重复组合的公式 : n! (r)! xr!

$\begin{aligned} \frac{23!}{(23 - 5)! \times 5!} \\ \frac{23 \times 22 \times 21 \times 20 \times 19}{5!} \\ \frac{4037880}{120} \\ 33, 649 p o s s i b l e t e a m s \end{aligned}$

::23! (23- 5) 3x5! 23x22x21x20x195! 403788012033,649 可能的小组

Example 2
::例2

How many three-color combinations can be selected from the seven-color basic rainbow?
::从七色基本彩虹中可以选取多少三色组合?

This is a $\begin{aligned} _{7} C_{3} \end{aligned}$ calculation, repeats allowed. $\begin{aligned} n = 7, r = 3 \end{aligned}$ , use the formula for repeating combinations: $\begin{aligned} \frac{(n + r - 1)!}{(n - 1)! \times r!} \end{aligned}$
::这是 7C3 计算, 重复允许。 n= 7, r= 3, 使用重复组合的公式 : (n+r-1) ! (n-1)! xr!

$\begin{aligned} \frac{(7 + 3 - 1)!}{(7 - 1)! \times 3!} \\ \frac{9!}{6! \times 3!} \\ \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \\ \frac{504}{6} \\ 84 c o l o r c o m b i n a t i o n s \end{aligned}$

:7+3- 1) ! (7- 1) ! x3! 9! 6! x3! 9x8x73x2x1504684 彩色组合

Example 3
::例3

How should $\begin{aligned} _{8} C_{2} \end{aligned}$ be read? How many combinations does it indicate?
::如何读8C2?它表示有多少组合?

$\begin{aligned} _{8} C_{2} \end{aligned}$ is read as “Choose 2 items at a time from 8 possibilities”
::8C2改为“从8种可能性中选择2项”

To calculate the number of combinations indicated, note that $\begin{aligned} n = 8 \end{aligned}$ and $\begin{aligned} r = 2 \end{aligned}$ . There are two possible solutions, depending on whether the situation allows for repeats.
::为计算所示组合的数量,请注意 n=8 和 r=2 。根据情况是否允许重复,有两种可能的解决办法。

If repeats are allowed, we use $\begin{aligned} \frac{(n + r - 1)!}{(n - 1)! \times r!} \end{aligned}$ , and we get: $\begin{aligned} \frac{(8 + 2 - 1)!}{(8 - 1)! \times 2!} = \frac{9!}{7! \times 2!} = \frac{9 \times 8}{2} = 36 \end{aligned}$
::如果允许重复, 我们使用 (n+r- 1) !! xr! ! ! ! ! ! ! !, 我们得到 : (8+2- 1) ! (8- 1) ! x2! =9! 7! x2! =9x82=36

If repeats are not allowed, use $\begin{aligned} \frac{n!}{(n - r)! \times r!} \end{aligned}$ , to get: $\begin{aligned} \frac{8!}{(8 - 2)! \times 2!} = \frac{8!}{6! \times 2!} = \frac{8 \times 7}{2} = 28 \end{aligned}$
::如果不允许重复, 请使用 n! ! xr! ! 获取 : 8! (8) 2! x2! = 8! 6! x2! = 8! 6! x2! =8x72=28

Review
::回顾

For questions 1-10, calculate the number of combinations indicated by the combination notation, on odd-numbered questions, assume repeats are allowed , otherwise assume no repeats.
::对于问题1-10,计算组合符号显示的组合数,在奇数问题上,假设允许重复,否则假设不重复。

$\begin{aligned} _{5} C_{3} \end{aligned}$
::5C3 5C3

$\begin{aligned} _{7} C_{2} \end{aligned}$
::7C2 7C2

$\begin{aligned} _{8} C_{3} \end{aligned}$
::8C3 8C3

$\begin{aligned} _{13} C_{7} \end{aligned}$
::13C7 13C7

$\begin{aligned} _{12} C_{4} \end{aligned}$
::12C4 12C4

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

$\begin{aligned} _{6} C_{5} \end{aligned}$
::6C5 6C5

$\begin{aligned} _{8} C_{5} \end{aligned}$
::8C58C5

$\begin{aligned} _{11} C_{3} \end{aligned}$
::11C3 11C3

$\begin{aligned} _{11} C_{3} \end{aligned}$
::11C3 11C3

How many unique 6-player teams can be picked from a pool of 19 players?
::从19个球员的球池里选出多少个独特的6人球队?

You are part of an 8-person trivia team. When you compete in trivia tournaments, players are chosen to answer questions based on random numbers generated between 1 and 8. If a tournament consists of 20 questions, how many ways could the team members be chosen to answer trivia questions?
::你是一个8人三维队的一部分。当你参加三维队比赛时,球员被选中根据1至8之间的随机数字回答问题。如果比赛由20个问题组成,那么可以选择多少方法回答三维队的问题?

You work in an ice cream shop. The most popular cone in the store is the “Monster Bellyache”, a massive four-scoop ice-cream bowl. If the shop carries seventeen flavors, how many unique “Monster Bellyaches” are there?
::你在一个冰淇淋店工作,商店里最受欢迎的甜甜圈是“贝利阿奇怪物”,一个大型的四勺冰淇淋碗,如果商店有十七种口味,有多少独一无二的“贝利阿奇怪物”?

The same ice cream shop carries three other options for ice-cream bowls: “Big Bellyache” (3 scoops), “Brain Freeze” (2-scoops), and “Diet Denter” (1-scoop). How many unique bowls of ice cream are possible, counting all four types of bowls?
::同样的冰淇淋店还有另外三种冰淇淋碗选择:“大贝利阿奇 ” ( 3 勺子 ) 、 “ 冻结” ( 2 勺子 ) 、 “ Diet Denter ” ( 1 勺子 ) 。有多少个独特的冰淇淋碗是可能的,包括所有四种碗?

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。

版权所有 © 2024 由瓜瓜科技thewaytolearn.cn设计。保留所有权利。