From a to b, you can take the train, bus or ship. In one day, there are four trains, two buses and three ships. How many different ways are there to get from A to B in a day by these means of transportation?

Because there are four modes of travel in a day-trains, cars and ships-and each mode can get from A to B, there are 423 = 9 different ways to get from A to B through these means of transportation.

Generally speaking, there are the following principles:

Addition principle: There are n ways to do something. Finish it. The first mode has m 1 different modes, the second mode has m2 different modes, …, and the nth mode has mn different modes. Then there are different ways for n = M 1 m2 10 mn to accomplish it.

(2) Let's look at the following questions:

There are three roads from village A to village B, and two roads from village B to village C. How many different roads are there from village A to village C via village B?

Here, there are three different ways from village A to village B, and after each of these three ways arrives at village B, there are two different ways from village B to village C. Therefore, there are 3X2=6 different ways from village A to village C via village B. 。

Generally speaking, there are the following principles:

Multiplication principle: to do one thing, it needs to be divided into n steps. There are m 1 different ways to do the first step, m2 different ways to do the second step, …, and mn different ways to do the n step. Then there are n = m 65438+m 2 … Mn different ways to do it.

There are 6 different math books on the upper shelf and 5 different Chinese books on the lower shelf.

1) How many different ways are there to get one of them?

2) How many ways to take a math book and a Chinese book?

Solution: (1) There are two ways to take books from the bookshelf: the first way is to take a math book from the upper level, and you can take any one of the six books. There are six ways; The second method is to get Chinese books from the lower level. You can choose any one of the five books. There are five ways. According to addition principle, the different numbers of Chinese books are 65 = 1 1.

Answer: There are 1 1 different ways to get a book from the shelf.

(2) Taking a math book and a Chinese book from the bookshelf can be completed in two steps: step one, there are six ways to take a math book; Step two, take a Chinese book. There are five ways. According to the principle of multiplication, the number of different methods is n = 6x5 = 30.

Answer: There are 30 different ways to get a math book and a Chinese book from the bookshelf.

Exercise: A classmate has four different ancient coins of Ming Dynasty and six different ancient coins of Qing Dynasty.

1) How many different ways are there to get one of them? 2) How many different ways are there to take an ancient Ming and Qing coin?

Example 2: (1) How many numbers can the numbers l, 2, 3, 4 and 5 form? How many digits are allowed to repeat three digits?

(2) How many numbers can the numbers L, 2, 3, 4 and 5 make up? Duplicate three digits are not allowed.

(3) How many numbers can the numbers 0, L, 2, 3, 4 and 5 form? Three digits are not allowed to be repeated?

Solution: Forming a three-digit number can be completed in three steps: the first step is to determine the number on the hundredth digit and choose a number from five. There are five ways to choose * * *; The second step is to determine the number on the decimal digit. Because the numbers are allowed to be duplicated,

There are five other options. The third step is to determine the number of units. Again, there are five options. According to the principle of multiplication, the three digits that can be composed are n = 5x5x5 = 125.

A: It can be composed of three digits: 125.

grade

Review basic principles

1. addition principle can do one thing in n ways. In the first way, there are m 1 different ways, in the second way, there are m2 different ways ... and in the nth way, there are mn different ways, so there are * * * ways to complete it.

N=m 1+m2+m3+…mn

Different methods.

2. multiplication principle: to do one thing, it needs to be divided into n steps. There are m 1 different ways to do the first step, m2 different ways to do the second step, ... and mn different ways to do the n step, so there are * * * ways to do it.

N=m 1？ m2？ m3？ …? manganese

Different methods.

3. The difference between the two principles:

Exercise 1

1. How many different kinds of air tickets do I need to prepare for direct flights from Beijing, Shanghai and Guangzhou?

2. How many non-repeating numbers can the numbers 1, 2 and 3 make up? Please list them one by one.

basic concept

1. What is an arrangement? From n different elements, any m () elements (the selected elements here are different) are arranged in a column in a certain order, which is called the arrangement of m elements from n different elements.

2. What do you mean by different arrangements? There is at least one difference between elements and order.

3. What is the same arrangement? Elements are arranged in the same order.

4. What is an agreement?

Examples and exercises

1. How many three digits can the number 1, 2, 3, 4 form?

2. Know the four elements A, B, C and D, ① Write out all the permutations of three elements at a time; Write out all the permutations of four elements at once.

number of permutations

1. Definition: All permutation numbers of m () elements from n different elements are called permutation numbers of m elements from n elements, which are represented by symbols.

Symbolize the number of permutations in the above questions.

2. Formula of permutation number: = n (n-1) (n-2) ... (n-m+1)

arrange

Process:

Review: (Guide students to review and sort out what they learned last class)

The definition of 1. permutation and some problems that should be paid attention to when understanding the definition of permutation;

2. Definition and calculation formula of permutation number

Or (where m≤n m, n? z)

3. The meaning of complete permutation and factorial; Rule 0! = 1

4. The application of the ideas of "classification" and "step by step" in the arrangement problem.

Second, the new grant:

Example 1: (1) Seven students stand in a row. How many different arrangements are there?

Solution: The problem can be regarded as: the complete arrangement of seven elements -= 5040.

(2) Seven students stand in two rows (top three, last four). * * * How many different arrangements are there?

Solution: According to the principle of stepwise counting: 7× 6× 5× 4× 3× 2× 1 = 7! =5040

(3) Seven students stand in a row with one in the middle. How many different arrangements are there?

Solution: The problem can be regarded as: the complete arrangement of the remaining six elements -= 720.

(4) Seven students stand in a row, and A and B can only stand at both ends. How many arrangements are there?

Solution: According to the principle of step-by-step counting: Step 1: A, bilibili is at both ends; The second step is to arrange all the remaining five students, and there are * * * =240 arrangement methods.

Five] Seven students stand in a row. How many ways can A and B not stand at the head and tail?

Scheme 1 (direct method): Step 1, choose two students from the remaining five students (except A and B) to stand in the front and back. The second step is to select five students from the remaining five students to arrange (completely arrange), so there are = 2400 arrangements.

Solution 2: (exclusion method) If A stands at the front, there is a method; If bilibili is at the end, there is a way; If A stands at the top and bilibili is at the bottom, there are two ways. So A can't reach the end, and bilibili can't reach the end. There are -+= 2400 ways.

Summary 1: For the problem of "being" and "not being", the "direct method" or "exclusion method" is often adopted, and some special elements can be given priority.

Exodus 2: Seven students stand in a row.

(1) How many adjacent permutations must students A and B have?

Solution: There is a way to "bind" two students A and B together as one element and arrange them all together with the other five elements (classmates). There is a way to "untie" students A and B and arrange them well. So this arrangement has = 1440.

(2) Students A, B and C are all adjacent. How many arrangements are there?

Solution: The method is the same as above, with a * * * = 720 kinds.

(3) Two students, A and B, must be adjacent. How many ways does C not stand at the head and tail?

Solution 1: Students A and B are "tied" together as an element. At this time, a * * * has six elements. Because c can't stand at the head and tail, we can choose two elements from the other five elements to put at the head and tail. There is a way; There is a way to arrange all the remaining four elements; Finally, there are ways to "untie" students A and B, so there are = 960 ways to arrange them.

Solution 2: Two students A and B are "tied" together as an element. At this time, a * * * has six elements. If C stands at the head or tail, there are two ways, then C does not stand at the head and tail, there is one way.

Solution 3: Tie students A and B together as an element. At this time, a * * * has six elements. Because C can't stand in the front row and the back row, you can choose a method from the other four positions, then arrange all the remaining five elements, and finally "unbind" students A and B. This arrangement is one.

Summary 2: For adjacent problems, the "binding method" is often adopted (binding first and then loosening).

Exodus 3: Seven students stand in a row.

(1) How many arrangements are there for students A and B not to be adjacent?

Solution 1: (exclusion method)

Solution 2: (Interpolation) One method is to arrange the other five students first, at this time they leave six positions (called "empty"), and then insert students A and B into these six positions (empty) respectively, so there is a method.

(2) How many arrangements are there for students A, B and C who cannot be adjacent to each other?

Solution: First, arrange the other four students in one way. At this time, they left five "blanks", and then inserted three students A, B and C into these five "blanks" in a way, so a * * * has = 1440.

Summary 3: For non-adjacent problems, interpolation method is often used (special elements will be considered later).

Third, summary:

1. For the constrained permutation problem, we should pay attention to the following types:

(1) Some elements cannot or must be arranged in a certain position;

(2) Some elements require even lines (that is, they must be adjacent);

(3) Some elements require separation (that is, they cannot be adjacent);

2. The basic method to solve the problem:

(1) There is a problem with the arrangement of special elements or special positions. Usually, special elements or special positions are arranged first, which is called the special element (position) priority method (optimal limit method);

(2) When some elements are required to be adjacent, they can be regarded as one element first, and then the internal arrangement of adjacent elements can be considered after being arranged with other elements. This method is called "binding method";

(3) When some elements are not adjacently arranged, you can arrange other elements first, and then insert these nonadjacent elements into the middle position. This method is called "interpolation";

⑷ When dealing with permutation problems, we can generally adopt direct and indirect thinking modes to seek effective problem-solving methods, which is the basis for learning permutation problems well.

4. Homework: "Arrange class hours 1-3" in "Classroom Practice"

Theme: Simple Application of Arrangement (2)

Objective: To enable students to learn how to calculate and solve simple practical problems with permutation number formula, further cultivate their ability to analyze and solve problems, and at the same time let students learn how to solve multiple problems.

Process:

First, let's review:

The definitions of 1. permutation and permutation number, and two formulas for calculating permutation number;

2. Three common queuing problems:

(1) Some elements cannot or must be arranged in a certain position-optimal limit method;

(2) Some elements require juxtaposition (that is, they must be adjacent)-binding method;

(3) Some elements require separation (that is, they cannot be adjacent)-interpolation.

3. The application of the idea of classified distribution.

Second, the new grant:

Example 1: arrange 6 out of 65,438+00 different cultural programs into a program list. If jane doe's solo program must not be ranked as the second program, how many different arrangements are there?

Option 1: (Considering from a special location)

Solution 2: (considering special elements) If you choose; If you don't choose:

Then * * * has += 136080.

Option 3: (indirect method) 136080

Example 2:

(1) Eight people are arranged in two rows, with four people in each row. Among them, Party A and Party B are in the front row and Party C is in the back row. How many different arrangements are there?

Briefly explain: A and B are in the front row; C is in the back row; All the rest are arranged.

So a * * * has = 5760 methods.

⑵ Five different commodities are arranged in a row on the shelf, among which two commodities A and B must be arranged together, and two commodities C and D are not arranged together. How many different arrangements are there?

Simple solution: (comprehensive use of "binding method" and "inserting empty method") A and B are tied together and arranged with E;

At this time, three vacancies are left, and two commodities, C and D, are placed in one * * *; Finally, A and B are "unbound", so a * * * has = 24 methods.

(3) Six movie tickets with the same serial number are distributed to three teachers and three students. If teachers and students are required to sit alternately, how many different ways are there?

Brief description: (classification) If the first one is a teacher, yes; If the first one is a student, yes.

So a * * * has 2 = 72 methods.

Example 3:

How many positive integers can (1) 1, 2, 3, 4, 5 be formed without repetition?

Brief description:

⑵ 1, 2, 3, 4, 5, which have no duplicates and are greater than 13 000, how many positive integers can be formed?

Scheme 1: Divided into two categories. One is that when the first digit is 1, the ten digits must be greater than or equal to 3. The other is that the first digit is not 1, so there is a way. So * * * number is greater than 13 000.

Solution 2: (Exclusion method) has a positive integer less than 13 000, so there is a positive integer greater than 13 000 = 1 14.

Example 4: Use 1, 3, 6, 7, 8, 9 to form non-repeating four digits, arranged from small to large.

What's the number of (1) 1 14? What's the date of 3 796?

Solution: (1) Because the thousands are 1, there are four * * *, so the thousands of 1 14 should be "3" and the ten digits are "1", that is, "3/kloc-0". Similarly, 12 numbers start with "36", "37" and "38" respectively, so the first two digits of the number 1 14 must be "39" and "3 968" ranks sixth, so "3 968" is 6548.

(2) As can be seen from the above, the number starting with "37" is preceded by 60+ 12+ 12 = 84, and 3796 ranks 1 1 (the second last) among the four numbers starting with "37", so it is 3796.

Example 5: Use 0, 1, 2, 3, 4, 5 to form a non-repetitive four-digit number, in which

(1) How many numbers are divisible by 25?

(2) How many digits are larger than single digits?

Solution: (1) The last two digits of four digits divisible by 25 can only be 25 and 50. There are four numbers ending in 50 and 25, so a * * * has += 2 1.

Note: The last two digits of four digits divisible by 25 can only be 25, 50, 75, 00.

(2) Use 0, 1, 2, 3, 4, 5 to form a four-digit number, which is not repeated, and each * * * has one. Because the relationship between ten digits and single digits is "equally possible" among the 300 numbers, one of the ten digits is larger than the single digits.

Combination (1)

1. Question:

Example 1: Choose two students from A, B and C to participate in activities on a certain day, of which 1 students participate in activities in the morning and 1 students participate in activities in the afternoon. How many different ways are there?

Example 2: Choose two students from A, B and C to take part in an activity. How many different ways are there?

Guided observation: Example 1 requires not only the selection of two students, but also "arrangement" in a certain order, while Example 2 only requires the selection of two students, regardless of the order.

Guiding topic: combination problem.

Second, the new grant:

The concept of 1. combination: Generally speaking, taking out m(m≤n) elements from n different elements for grouping is called taking out the combination of m elements from n different elements.

Note: 1. Different element 2. "Just take it without excluding it"-Symptom 3. The same combination: the same elements.

Determine which of the following problems is a permutation problem and which is a combination problem:

(1) Choose two scenic spots from A, B, C and D for sightseeing; (combination)

(2) Choose two students from A, B, C and D as monitor and secretary of the Youth League branch. (arrangement)

2. Concept of combination number: The number of all combinations of m(m≤n) elements taken from n different elements is called the combination number of m elements taken from n different elements, which is expressed by symbols.

For example, the combination of two students selected from three students in Example 2 can be: A, B, C, B, C, that is, there is one combination.

Another example: choose two combinations from AB, AC, AD, BC, BD and CD for sightseeing, namely:

When explaining, students must analyze whether the problem to be solved is arrangement or combination, depending on whether it is related to the order. Then how to calculate?

3. Derivation of combination number formula

(1) Question: What is the combination number of three elements of four different elements: A, B, C and D?

Revelation: Because the arrangement is first combined and then arranged, we can get the arrangement number of three elements in four different elements, so we can examine the relationship between and, as follows:

Combination arrangement

It can be seen that each combination corresponds to six different arrangements. Therefore, finding the permutation number of three elements from four different elements can be divided into the following two steps: ① consider taking out the combination of three elements from four different elements, and * * * has one; ② The three different elements of each combination have different arrangements, which is obtained from the principle of step-by-step counting: =, so:.

⑵ Summary: Generally speaking, finding the arrangement number of M elements from N different elements can be divided into the following two steps: ① finding the combination number of M elements from N different elements; ② Find the total arrangement number of M elements in each combination, and according to the principle of distribution counting, we get: =

(3) The formula of combination number:

or

4. Comment on examples

Example 1. 6 Different books are distributed to students A, B and C, and each student gets two books. How many different grades are there?

Law?

Brief description:

Example 2.4 Boys and 6 girls form a three-person practical activity group with at least/kloc-0 boys. How many ways are there?

Scheme 1: (Direct method) There are three groups, namely: 3 males, 2 males, 1 female, 1 male and 2 females, so one * * has ++= 100 methods.

Solution 2: (Indirect Method)

In addition, when solving practical problems, we should first look at whether it is related to order, so as to determine whether it is a permutation problem or a combination problem. If necessary, we should use the principles of classification and gradual counting.

Combination (2)

Process:

First, review review:

1. Review the related contents of permutation and combination:

Key points: arrangement order; Combination disorder.

Second, the new grant:

Attribute of 1. Combination number 1:.

Understanding: Generally speaking, after taking out M elements from N different elements, n- m elements remain.

In order to extract every combination of M elements from N different elements, corresponding to every combination of the remaining n- m elements, the combination number of M elements of N different elements is equal to the combination number of n- m elements of N elements, that is, here, we mainly embody the idea that "taking method" and "keeping method" are "one-to-one correspondence".

Evidence:

Once again Ⅷ

Note: 1 We stipulate that

Characteristics of the equation: the subscripts on both sides of the equation are the same, and the sum of the subscripts is equal to the subscripts.

The function of this attribute: at this time, the calculation can be changed to calculation, which can simplify the operation.

For example: = = = 2002.

4 or

2. Example 1: (Textbook 10 1 Example 4) A pocket contains seven white balls and 1 black balls of the same size.

(1) Take three balls out of your pocket. How many ways are there?

(2) Take out three balls from the pocket to make them contain 1 black balls. How many ways are there?

⑶ Take three balls out of your pocket. How many ways can you make them free of black balls?

Solution: (1) (2) (3)

Guide students to discover: Why?

We can explain it this way: the three balls taken out of the eight balls in the pocket can be divided into two categories: one category contains 1 black balls, and the other category does not contain black balls. Therefore, according to the principle of classified counting, the above equation holds.

Generally speaking, the number of combinations of m elements is extracted from these n+ 1 different elements, that is, these combinations can be divided into two categories: one contains elements and the other does not. The combination containing m-1 elements is composed of m- 1 elements, and * * has one; An excluded combination consists of m of the n elements, with * * *. According to the principle of classified counting, another property of combination number can be obtained. Here, it mainly embodies the inductive thought from special to general, that is, the classification thought of "including and excluding its elements"

First, knowledge review:

1. Review the related contents of permutation and combination:

Or emphasize: arrangement-order; Combination disorder.

2. Formulas and related properties of permutation number and combination number

Attribute 1: Attribute 2: =+

Common equations:

3. Exercise: Deal with 76 examples of teaching and testing.

Second, the example notes:

Of the 1. 100 products, 90 are qualified products and 4 are defective products 10. Now, we have selected them for inspection.

(1) How many ways can you make them all defect-free?

⑵ How many ways can we get at least 1 defective products?

(3) How many ways can we adopt incomplete defective products?

Solution: (1);

⑵ ;

⑶ .

Example 2. Take five balls from * *10, 1 10, and make the sum of the five balls odd, then one * * * has.

Solution: divided into three categories: 1 odd 4 pairs; 3 odd numbers and 2 accidents; 5 odd 1 occasionally

So a * * * has a++.

Example 3. Eight young people, five of whom are qualified as English translators; Four young people are qualified to be German translators.

Translation work (including 1 a young man who has the best of both worlds), five young people are selected to undertake a task, of which three are engaged in English translation and two are engaged in German translation. How many different ways are there?

Solution: We can be divided into three categories:

(1) Let young people who can have both be engaged in English translation, including;

(2) Let young people who can hold two jobs engage in German translation, including:

Let young people who can do both jobs do nothing, ok.

So a * * * has ++= 42 methods.

Example 4. Party A, Party B and Party C are on duty from Monday to Saturday. Party A is not worth Monday and Party B is not worth Saturday. How many different duty tables can be arranged?

Solution 1: (exclusion method)

Option 2: it is divided into two categories: one is that A is not worth Monday or Saturday, and there is; The other is that A is not worth Monday, but it is worth Saturday, so a * * * has += 42 methods.

Example 5.6 All the different books are given to five people, and each person has at least 1 book. How many different ways are there to give books?

Solution: The first step is to take any two books from six different books and "bundle" them together as an element. The second step is to divide the five "different books" into five people. According to the principle of step-by-step counting, there are = 1800 methods for a * * *.

Change the title 1: 6 All the different books for 5 people. How many different ways are there to send books?

Change 2: 5 Give all the different books to 6 people, each with a maximum of 1 book. How many different ways are there to send books?

Change 3: 5 All the same books are given to 6 people, each with a maximum of 1 book. How many different ways are there to send books?

Answer: 1. 2.； three ..

3. Summary: 1. The definition of combination, the formula of combination number and its two properties;

2. Combined application: distinguish whether to sort.

Fourth, homework: "3+X" combination basic training

Class practice class 10 combination 4

Combination (4)

Subject: Comprehensive application of combination and combination number (2)

Objective: To have a systematic understanding of the knowledge of permutation and combination, master some common questions and solving methods of permutation and combination, and be able to solve permutation and combination problems by using two principles and permutation and combination concepts.

Process:

First, knowledge review:

1. two basic principles;

2. Related concepts and properties of permutation and combination.

Second, the example notes:

Example 1.6 according to the following requirements, how many different selection methods are there for different books:

(1) Give two copies to Party A, Party B and Party C respectively;

(2) Divided into three parts, each with two copies;

(3) Divided into three parts, one part, two parts and three parts;

(4) Distribute to Party A, Party B and Party C, with one copy for each, two copies for each and three copies for each;

5] Give it to Party A, Party B and Party C, each with at least one copy.

Solution: (1) According to the principle of step-by-step counting, there are:

(2) Assign A, B and C in two ways. This process can be completed in two steps: the first step is divided into three parts, each of which has two methods; Second, there is a way to distribute these three copies to students A, B and C. According to the principle of step-by-step counting, we can get:, so. So there are 15 ways to divide them into three parts, two for each part.

Note: This question is about "even grouping" in grouping.

(3) This is a problem of uneven grouping, and there is a way.

(4) On the basis of (3), the whole arrangement is being made, so there is a way.

5] It can be divided into three types: ① "type 2, 2, 2", that is, the distribution in ① has methods; (2) "1,2,3 type", that is, the distribution in (4), there are methods; ③ "1,1,4 type", there is a method. So a * * has 90+360+90 = 540 methods.

Example 2. Seven athletes with different heights stand in a row. A, B and C are arranged from left to right and from high to low, and they are not adjacent to each other.

Solution: (interpolation method) One way is to arrange all the remaining four students, and then another way is to insert three students A, B and C into five empty positions (but there is no need to arrange them). According to the principle of step-by-step counting, there are = 240 ways in one * *.

Example 3. (1) Put four different balls in four different boxes. How many different ways are there to throw a ball?

How many ways are there to put four different balls in four different boxes with an empty box?

Solution: (1) According to the principle of step-by-step counting, there are methods.

(2) (Binding method) Step 1, bind any two balls of four different balls together as an element. Step 2: Take three balls from four different boxes and put them into a method. So a * * has = 144 methods.

Example 4. There are ten street lamps numbered 1, 2, 3, …, 10 on the road. In order to save electricity without affecting lighting, three lights can be turned off, but two or three adjacent lights cannot be turned off at the same time. The lights at both ends can't be turned off. How many different ways are there to turn off the lights?

Solution: (Insertion method) This problem is equivalent to inserting three extinguished lamps in six gaps between seven lighted street lamps, so the total number of methods is one.

Example 5. The numbers 0, 1, 2, …, 8 are written on the nine cards respectively. Take out three cards and line them up to form a three-digit number. If 6 can be used as 9, how many three digits can be formed?

Solution: There are two kinds: ① If 6 is taken out, there is a way; (2) if 6 didn't take, there is a way. According to the principle of classified counting, there are += 602 methods for a * * *.