不论是否重复使用基本计算原则

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A lock has the digits 0-39. A series of three numbers unlocks the lock. How many possible unlocking combinations are there if the numbers cannot be repeated?
::锁有数字 0 - 39 。 三个数字序列可以解锁。 如果数字无法重复, 有多少组合可以解锁 ?

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Fundamental Counting Principle 
::基本计数原则

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Consider a phone number. A phone number consists entirely of numbers that might be repeated. In this concept we will look at how to determine the total number of possible combinations of items which may be repeated.
::考虑一个电话号码。电话号码完全包括可能重复的数字。在这个概念中,我们将研究如何确定可能重复的项目组合的总数。

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Let's solve the following problems.
::让我们解决以下的问题。

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1. A license plate consists of three letters and four numbers in the state of Virginia. If letters and numbers can be repeated, how many possible license plates can be made?
   ::在弗吉尼亚州,一个牌照由三个字母和四个号码组成,如果可以重复字母和号码,可以制作多少个可能的牌照?

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I f we consider the three slots for the letters, how many letters can be chosen to place in each slot? How about the four slots for the numbers? If there are no restrictions, i.e. letter and numbers can be repeated, the total number of license plates is:
::如果我们考虑字母的三个空档,每个空档可选用多少字母?数字的四个空档如何?如果没有限制,即可以重复字母和号码,牌照的总数是:

$$\begin{align*}\underline{{\color{red}26}} \times \underline{{\color{red}26}} \times \underline{{\color{red}26}} \times \underline{{\color{blue}10}} \times \underline{{\color{blue}10}} \times \underline{{\color{blue}10}} \times \underline{{\color{blue}10}}=175,760,000\end{align*}$$

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Now, what if letters or numbers could not be repeated? Well, after the first letter is chosen, how many letters could fill the next spot? Since we started with 26, there would be 25 unused letters for the second slot and 24 for the third slot. Similarly with the numbers, there would be one less each time:
::现在,如果无法重复字母或数字呢? 那么,在第一个字母被选中后, 下个位置能填满多少字母? 自从我们用26个字母开始, 第二个位置会有25个未用字母, 第三个位置会有24个未用字母。 和数字一样, 每次少一个字母:

$$\begin{align*}\underline{{\color{red}26}} \times \underline{{\color{red}25}} \times \underline{{\color{red}24}} \times \underline{{\color{blue}10}} \times \underline{{\color{blue}9}} \times \underline{{\color{blue}8}} \times \underline{{\color{blue}7}}=78,624,000\end{align*}$$

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2. How many unique five letter passwords can be made? How many can be made if no letter is to be repeated?
   ::可以作出多少个独特的五个字母密码?如果不重复字母,可以作出多少字母密码?

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Since there are 26 letters from which to choose for each of 5 slots, the number of unique passwords can be found by multiplying 26 by itself 5 times or $\begin{align*}(26)^5=11,881,376\end{align*}$ . If we do not repeat letters, then we need to subtract one each time we multiply: $\begin{align*}26 \times 25 \times 24 \times 23 \times 22=7,893,600\end{align*}$ .
::由于5个空档中每个空档都有26个字母可供选择,所以可以将26个密码本身乘以5倍或(265=11,881,376)来找到独特密码的数目。如果我们不重复字母,那么我们每次乘以:26x25x24x23x22=7,893,600,就需要减去一个。

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3. How many unique 4 digit numbers can be made? What if no digits can be repeated?
   ::有多少个独特的四位数数字可以做成?如果没有数字可以重复呢?如果没有数字可以重复呢?

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For the first part, consider that in order for the number to be a four digit number, the first digit cannot be zero. So, we start with only 9 digits for the first slot. The second slot could be filled with any of the ten digits and so on:
::第一部分,考虑一下,如果数字要为四位数,第一个数字不能为零。所以,我们从第一个空位的九个数字开始。第二个空位可以填满十个数字中的任何一个,等等:

$$\begin{align*}\underline{9} \times \underline{10} \times \underline{10} \times \underline{10}=9000.\end{align*}$$

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For the second part, in which digits cannot be repeated, we would still have 9 possible digits for the first slot, then we’d have 9 again for the second slot (we cannot repeat the first digit, but we can add 0 back into the mix), then 8 for the third slot and 7 for the final slot:
::在不能重复数字的第二部分,我们的第一个空档仍有9个可能的数字,然后第二个空档将有9个可能的数字(我们不能重复第一个数字,但我们可以在组合中再加0个数字),第三个空档还有8个数字,最后一个空档有7个数字:

$$\begin{align*}\underline{9} \times \underline{9} \times \underline{8} \times \underline{7}=4536.\end{align*}$$

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Examples
::实例

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Example 1
::例1

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Earlier, you were asked to find the number of  possible unlocking combinations if the numbers cannot be repeated. 
::早些时候,有人要求您在无法重复数字时寻找可能的解锁组合数 。

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Since there are 40 numbers from which to choose for each of 3 slots, the number of unique passwords can be found by multiplying 40 by itself 3 times or $\begin{align*}(40)^3=64,000\end{align*}$ . However, we cannot repeat numbers so we need to subtract one each time we multiply: $\begin{align*}40 \times 39 \times 38 =59,280\end{align*}$ .
::3个空档中,每个空档都有40个数字可供选择,因此,通过将40个密码本身乘以3倍或(403=64 000),可以找到独特的密码数。然而,我们不能重复数字,因此每次乘以40x39x38=59280,就需要减去一个密码数。

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Therefore, there are 59,280 possible unlocking combinations.
::因此,有59 280种可能解锁的组合。

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Example 2
::例2

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How many unique passwords can be made from 6 letters followed by 1 number or symbol if there are ten possible symbols? No letters or numbers can be repeated.
::如果可能有十个符号, 6个字母之后的1个编号或符号可以密码多少? 没有字母或数字可以重复 。

$\begin{align*}\underline{{\color{red}26}} \times \underline{{\color{red}25}} \times \underline{{\color{red}24}} \times \underline{{\color{red}23}} \times \underline{{\color{red}22}} \times \underline{{\color{red}21}} \times \underline{{\color{blue}20}}=3,315,312,000\end{align*}$

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Example 3
::例3

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If a license plate has three letters and three numbers, how many possible combinations can be made?
::如果牌照有3个字母和3个号码,可以组合多少个?

$\begin{align*}\underline{{\color{red}26}} \times \underline{{\color{red}26}} \times \underline{{\color{red}26}} \times \underline{{\color{blue}10}} \times \underline{{\color{blue}10}} \times \underline{{\color{blue}10}} =17,576,000\end{align*}$

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Example 4
::例4

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In a seven digit phone number, the first three digits represent the exchange. If, within a particular area code, there are 53 exchanges, how many phone numbers can be made
::在7位数的电话号码中,头3位数代表交换台。如果在特定区域代码内有53个交换台,那么可以拨打多少个电话号码。

$\begin{align*}\underline{{\color{red}53}} \times \underline{{\color{blue}10}} \times \underline{{\color{blue}10}} \times \underline{{\color{blue}10}} \times \underline{{\color{blue}10}}=530,000\end{align*}$

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Review
::回顾

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Use the Fundamental Counting Principle to answer the following questions. Refer back to the examples and guided practice for help.
::使用 " 基本计算原则 " 回答下列问题。 参考实例和引导做法寻求帮助。

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1. How many six digit numbers can be formed if no digits can be repeated?
   ::如果没有数字可以重复,可以形成多少六位数?
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2. How many five digit numbers can be formed that end in 5?
   ::5分钟内能形成多少5位数?
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3. How many license plates can be formed of 4 letters followed by 2 numbers?
   ::有多少牌照可以由4个字母组成,然后是2个号码?
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4. How many seven digit phone numbers can be made if there are 75 exchanges in the area?
   ::如果该地区有75个交换机,可以拨打多少七个数字电话号码?
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5. How many four letter pins (codes) can be made?
   ::可以做多少四个字母针(代码)?
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6. How many four number/letter pins can be made if no number or letter can be repeated?
   ::如果没有数字或字母可以重复,可以打多少个号码/字母记号?
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7. How many different ways can nine unique novels be arranged on a shelf?
   ::在书架上可以用多少种不同的方式 安排九本独特的小说?
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8. How many different three scoop cones can be made from 12 flavors of ice cream allowing for repetition? What if no flavors can be repeated?
   ::从12种冰淇淋的口味中可以制造出多少不同的三勺锥形冰淇淋,这样可以重复吗?如果没有调味品可以重复呢?
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9. How many different driver’s license numbers can be formed by 2 letters followed by 6 numbers?
   ::有多少不同的驾驶执照号码可以用两个字母和六个号码组成?
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10. How many student ID numbers can be made by 4 random digits (zero cannot come first) followed by the student’s grade (9, 10, 11 or 12). Example: 5422-12 for a $\begin{align*}12^{th}\end{align*}$ grader.
    ::有多少学生身份证号码可以用4位随机数(零位数不能先到),然后是学生的年级(9、10、11或12)。 例如:12年级学生5422-12。

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Review (Answers)
::回顾(答复)

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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 。