重复重复的变换

播放段落

If a bank account consists of eight digits between zero and nine, and repeated digits are allowed, how many possible account numbers are there?
::如果银行账户由8位数组成,介于0至9之间,允许重复数字,那么有多少个可能的账户号码?

播放段落

This is not the same question as it would be if repeats were not allowed. At the end of this lesson, we’ll return to this question and review the difference.
::这个问题与不允许重复的问题不同。 在本课结束时,我们将回到这个问题,并审视差异。

播放段落

Permutations with Repeats 
::重复重复的变换

播放段落

When we consider permutations, we often specify that repeated values are not allowed, simply because many real-world situations would not support the concept. For instance, if you are calculating the number of possible seating arrangements for six people, it would be pretty silly to include any possibilities with the same person in multiple seats (unless you have a cloning machine in the closet!).
::当我们考虑变换时,我们常常明确指出,不允许重复重复的价值观,仅仅因为许多现实世界局势不支持这一概念。 比如,如果计算六人可能的座位安排数量,那么把同一人放在多个座位上的可能性都包括在内(除非衣柜里有克隆机 ! ) 是很愚蠢的。

播放段落

That said, there are times when you do need to include duplicates, such as the bank account question in the concept section above. The number of possible permutations in a set can be much greater when repeated values are allowed, but the calculation is actually simpler in concept.
::尽管如此,有时你确实需要列入重复数据,如上文概念一节中的银行账户问题。 如果允许重复值,一组中可能的变差数量可能要大得多,但计算实际上在概念上比较简单。

播放段落

To calculate the number of possible permutations of  $\begin{align*}r\end{align*}$ items from $\begin{align*}n\end{align*}$ available items, simply raise  $\begin{align*}n\end{align*}$ to the power of $\begin{align*}r\end{align*}$ :
::为计算r项从 n 可用项中可能的变换次数,只需将 n 提高到 r 的功率 :

播放段落

$$\begin{align*}n^r: (number \ available)^{number \ chosen}\end{align*}$$
::nr: nr( 可用数目) 选定数目

播放段落

Real-World Application: Ice Cream 
::真实世界应用程序: 冰淇淋

播放段落

The ice cream shop on the corner carries 27 flavors of ice cream, how many different 4-scoop cones can be created there?
::街角的冰淇淋店 配有27种口味的冰淇淋 在那里能创造出多少不同的 4层锥形锥形形形形色色的冰淇淋?

播放段落

There are 27 flavors, so $\begin{align*}n=27\end{align*}$ . We are creating 4-scoop cones, so $\begin{align*}r=4\end{align*}$ .
::我们有27种口味,所以是27种。我们正在创造4层锥形锥形,所以r=4。

播放段落

$$\begin{align*}Number \ of \ unique \ cones=27^4=531,441\end{align*}$$
::唯一锥体数=274=531,441

播放段落

Real-World Application: Locker Combinations 
::Real- World 应用程序: 锁定组合

播放段落

Lockers in your school are each three digits 0-9. How many different combinations are possible?
::你们学校的装箱每3位数 0 -9 。 有多少不同的组合是可能的?

播放段落

There are ten digits, so $\begin{align*}n=10\end{align*}$ . We are looking for arrangements of three digits each, so $\begin{align*}r=3\end{align*}$ .
::有十位数, 所以 n=10。 我们正在寻找三个位数的安排, 所以 r=3 。

播放段落

$$\begin{align*}Number \ of \ locker \ combinations=10^3=1000\end{align*}$$
::储物柜组合数=103=1000

播放段落

Real-World Application: Trivia 
::真实世界应用:特里维亚

播放段落

Keith’s trivia challenge team competes in trivia competitions all over the U.S. During a competition, the members are numbered 1-6, and a die is cast before each of the twenty questions in the challenge to decide which team member must answer the question. How many possible ways are there for Keith’s team to step up to the podium and answer trivia questions in a single meet?
::基思的三维挑战团队在美国各地的三维竞赛中竞争。 在一场竞赛中,会员人数为1-6,在挑战中的20个问题中,每个问题都会死亡,以决定哪个团队成员必须回答这个问题。 基思的团队在一次会面中走上讲台并回答三维问题有多少可能的方法?

播放段落

Keith’s team has six members, and there are twenty questions. There are six possible choices for the first question, and six possible  $\begin{align*}2^{nd}\end{align*}$ choices for each of them, resulting in $\begin{align*}6^2=36\end{align*}$  possibilities for the first two questions. Since there are twenty questions, the total number of possible lineups is $\begin{align*}6^{20} \approx 3.656 \times 10^{15}\end{align*}$ .
::基思的团队有6名成员,共有20个问题。 第一个问题有6个可能的选项,每个问题有6个可能的选项,每个问题有6个可能的选项,导致头两个问题有62=36的可能性。 由于有20个问题,可能排队的总数是6203.656×1015。

播放段落

Earlier  Problem Revisited
::重审先前的问题

播放段落

If a bank account consists of eight digits between zero and nine, and repeated digits are allowed, how many possible account numbers are there?
::如果银行账户由8位数组成,介于0至9之间,允许重复数字,那么有多少个可能的账户号码?

播放段落

Each digit has ten possibilities, and there are eight digits:  $\begin{align*}10^8=100,000,000\end{align*}$ possible bank accounts.
::每个数字有10种可能性,有8种数字:108=10万个可能的银行账户。

播放段落

Examples 
::实例

播放段落

Example 1
::例1

播放段落

How many different ice cream cones can be made with four scoops of ice cream, if there are 12 flavors to choose from?
::如果有12种口味可供选择的话,用4勺冰淇淋可以做出多少不同的冰淇淋锥?

播放段落

Since duplication is allowed, the number of possible permutations can be calculated with the formula $\begin{align*}n^r\end{align*}$ :
::由于允许重复,可用公式 nr 计算可能的变异次数:

$$\begin{align*}12^4=20, 736\end{align*}$$

播放段落

Example 2
::例2

播放段落

How many permutations are possible with seven units composed of the digits 0-9, duplication allowed?
::7个单位由数字0-9组成,允许重复,可以进行多少次变换?

播放段落

There are ten digits to choose from, and we are making permutations of seven digits each:
::我们从中选择十位数, 我们各做七位数的变换:

$$\begin{align*}10^7 =10, 000, 000\end{align*}$$

播放段落

Example 3
::例3

播放段落

How many four-letter permutations are possible using the letters of the alphabet?
::使用字母字母字母有多少四个字母的字母排列是可能的?

播放段落

There are twenty-six letters, and we are building permutations of four letters each:
::共有26个字母,我们正在建立4个字母的变换,每个字母:

$$\begin{align*}26^4=456,976\end{align*}$$

播放段落

Review 
::回顾

播放段落

For questions 1-12, calculate the number of possible permutations, duplicate values are allowed.
::对于问题1-12, 计算可能的变异次数, 允许重复值 。

播放段落
1. Using five decks of cards, permutations of five cards each.
   ::使用五张牌牌 每人五张牌
播放段落
2. Using the letters A-G and numbers 1-5, arrangements of eight units each.
   ::使用字母A-G和编号1-5,各安排八个单位。
播放段落
3. Using the letters in the word “combine”.
   ::使用“combine”一词中的字母。
播放段落
4. Seven digit arrangements of the numbers 0-9.
   ::数字0 -9的七位数安排。
播放段落
5. Ice cream cones with three scoops chosen from 19 flavors.
   ::从19种口味中选了三勺冰淇淋甜筒
播放段落
6. 1 cent, 5 cent, 10 cent, and 25 cent coins, in arrangements of five coins at a time.
   ::一分、五分、10%和25分硬币,一次五分。
播放段落
7. Letters A-F, in arrangements of six letters each.
   ::字母A -F,各安排六个字母
播放段落
8. Roll a 10-sided die seven times.
   ::滚动一个十面形死亡七次。
播放段落
9. Roll a standard die five times.
   ::滚动标准死亡5次。
播放段落
10. How many locker combinations are possible using three digits on a ten-digit dial?
    ::在10位数的拨号上使用3位数 有多少个储物柜组合组合?
播放段落
11. How many unique passwords can be made from the letters of the word “remix”?
    ::从“remix”一词的字母中可以输入多少个独特的密码?
播放段落
12. How many unique passwords can be made from the letters in “portable”?
    ::从这些“便携式”信件中可以打出多少个独特的密码?

播放段落

Review (Answers)
::回顾(答复)

播放段落

To view the Review answers, open this and look for section 8.3. 
::查看审查的答案,打开这一答案,并查看第8.3节。