C5 takes 3 = (5× 4× 3)/(3× 2× 1) C6 takes 2 = (6× 5)/(2× 1).

From these two examples, we can see that

The formula for CM to take n is a hierarchy in which the product of seed number m and its own continuous n natural numbers is taken as the numerator and n as the denominator.

P53=5×4×3 P66=6×5×4×3×2× 1

Through these two examples

The product of PMN = = M and its own continuous n natural numbers. When N = M, it is the hierarchy of m.

The essence of permutation and combination is to study "the possibilities of ordered and disordered arrangement of m (m≤n) elements" from n different elements. The signs that distinguish permutation and combination are "order" and "disorder".

There are two modes of thinking to solve the problem of permutation and combination:

One is to see whether the problem is orderly or disorderly. Orderly use "arrangement" and disorderly use "combination";

The second is to see whether the problem needs to be classified or step by step. Use "addition" for classification and "multiplication" for step by step.

Classification: "There are n ways to do one thing and complete it" is the classification of all the ways to complete it. When classifying, we should first determine a suitable classification standard according to the characteristics of the problem, and then classify it under this standard; Secondly, we should pay attention to two basic principles when classifying: ① Any method to accomplish this must belong to a certain category; ② Two methods belonging to two different categories are different methods.

Step by step: "To do one thing, it needs to be divided into n steps", which means that any way to accomplish it must be divided into n steps. In step by step, we should first determine a feasible step by step standard according to the characteristics of the problem; Secondly, the steps should be set to meet the needs of completing this matter. Only by continuously completing these N steps can this matter be finally completed.

The difference between the two principles is that one is related to classification and the other is related to gradual progress. If there are n ways to accomplish a thing, these n ways are independent of each other, and either way can be accomplished alone, addition principle is used to find the number of ways to accomplish it. If a thing needs to be divided into n steps, which are indispensable, that is, all the steps need to be completed in turn, and there are several different ways to complete each step, then use the principle of multiplication to find the way to complete it.

When solving the application problems of permutation and combination, we should pay attention to the following points:

1. Common propositional forms of permutation problems with restrictions:

"Yes" and "No"

"adjacent" and "not adjacent"

When solving problems, we should master the basic ideas and methods of solving problems;

⑴ The "combined cell method" is often used to solve the problem of "adjacency", and two or more cells can be regarded as one cell, which is the most commonly used method to deal with adjacency.

⑵ Interpolation arrangement is the most commonly used method to solve the problem of "non-adjacency".

⑶ The question of "being" and "not being" often involves special elements or special positions, which are usually arranged first.

(4) For the arrangement of elements with order restriction, the order restriction can be ignored first, and the results can be obtained by using the facts of the specified order after the arrangement is completed.

2. Conditional combination problem, common proposition form:

"contain" and "not contain"

"minimum" and "maximum"

The common methods to solve problems are direct method or indirect method.

3. When dealing with the comprehensive problem of permutation and combination, it is the most basic and important thinking method to solve the problem of permutation and combination by analyzing the conditions, classifying according to the nature of the elements, and applying the two principles step by step according to the occurrence process of the event.

Provide 10 exercises for everyone to practice.

1, three sides are integers, and the number of triangles with the maximum side length of 1 1 is (c).

25 (B)26 (C)36 (D)37。

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analyse

According to the principle of triangular sides, the sum of the two sides is greater than the third side, and the difference between the two sides is smaller than the third side.

The maximum visible edge is 1 1.

Then the sum of the two outer sides cannot exceed 22, because when all three sides are 1 1, the sum of the two sides is the largest.

So let's start with the length of one side.

If it is 1 1, then the length of the other side is 1 1, 1 0,9,8,7,6, ...1.

If it is 10, the length of the other side is 10, 9, 8.2,

(it can't be 1, otherwise the sum of them is less than 1 1, and it can't be 1 1, because the first case includes the combination of1,10).

If it is 9, the length of the other side is 9, 8, 7, ... 3.

(The reason is the same as above, indicating that the law has appeared.)

The total frequency of regularity is11+9+7+.1= (1+1) × 6 ÷ 2 = 36.

2、

(1) How many different ways are there to put four letters into three mailboxes?

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There are three options to analyze each letter. The relationship between letters is step by step. For example, if I put the letter 1 first, there are three possibilities. Then I put the second letter until the fourth letter. There are three possibilities, so the step relation belongs to the multiplication principle, that is, 3× 3× 3 = 3 4.

(2) How many different accommodation methods are there for three passengers staying in four hotels?

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Similar to the above situation, each of our passengers has four choices. There is no relationship between choices, no classification. It is a gradual relationship. For example, we arrange four choices for the first passenger and then four choices for the second passenger. Knowing the last passenger is also four possibilities. According to the step-by-step principle, it belongs to the multiplication relation, that is, 4× 4 = 4 3.

(3) Eight different books, three for three students, one for each. How many different ways are there?

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Analyze and do step by step.

Step 1: We choose three books, that is, how many possibilities C8 takes 3 = 56.

Step 2: Distribute to 3 students. P33 = 6 species.

Here is a brief introduction to why it is P33. Let's see if the first student can choose three books. After the selection is completed, the second student has only two choices, and the last student has no choice. That is, 3×2× 1. This is a step-by-step selection that conforms to the principle of multiplication. The most common example is 1, 2, 3, 4. How many four digits can these four numbers make up? It also satisfies such a principle of gradual progress. P is used to calculate because there are constraints between each step, that is, the choice of the next step is compressed by the previous step.

So the result of this problem is 56× 6 = 336.

3、

Seven students lined up in a horizontal row to take pictures.

(1)A How many different ways are there to stop at the head and the tail? (3600)

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analyse

We finished the problem in two steps.

Step 1: First, select C5 as one of the five positions where Row A should be in the middle, that is, 1 = 5.

Step 2: The remaining 6 people meet the P principle P66=720.

So the total is 720× 5 = 3600.

(2) How many different ways can a B only be ranked first or last? ( 1440)

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analyse

Step 1: Determine the place where B takes the lead, and choose C2 to take 1 = 2.

Step 2: The remaining 6 people meet the P principle P66=720.

Then the total is 720× 2 = 1440.

(3) When A is not at the head or tail and B is not in the middle, how many different arrangements are there? (3 120)

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Analyze special circumstances and arrange special circumstances first.

The first case: A is not at the beginning of the line or in the middle.

Remove three positions, leaving four positions for Party A to choose C4 1 = 4, and the remaining six positions are placed in the middle position, that is, except Party A and Party B, all five people can start with 5, and the remaining five positions meet the P principle, that is, 5× p55 = 5× 120 = 600, and the total number is 4× 600 = 2400.

Situation 2: A is not in the front row, but in the middle position.

Then the remaining six positions satisfy P66=720.

Because it is classified discussion, the final result is the sum of two situations, that is, 2400+720 = 3 120.

(4) How many permutations must A and B be adjacent? ( 1440)

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Analyze the binding principle of adjacent use, that is, two people become one person and seven bits become six, that is, step by step discussion.

No. 1: select position C6 and take 1 = 6.

No.2: The two selected positions are aligned with Party A and Party B, that is, P22 = 2.

Then the number of species meeting the requirements of both parties is 2× 6 = 12.

The other five people meet the rule of P55 = 120.

The final result is120×12 =1440.

(5) How many different arrangements must A have on the left (not necessarily adjacent) of B? (2520)

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analyse

This topic is very good. No matter how they are arranged, the probability that A appears on the left side of B is the same as that on the right side of B, so if the left and right problems are not considered, the total number is P77=5040. According to the principle of equal left and right probabilities, the number of species ranked on the left side is 5040 ÷ 2 = 2520.

4. Use the numbers 0, 1, 2, 3, 4, 5 to form a number without repeated numbers.

How many four digits can (1) form? (300)

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There are only five possibilities to analyze four digits from high to low and then to high, and they cannot be ranked as 0.

The next three positions satisfy the P53 principle = 5× 4× 3 = 60, that is, the total number is 60× 5 = 300.

(2) How many natural numbers can be formed? ( 163 1)

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Analysis of natural numbers starts with single digits.

Subsituation

1 digit: C6 takes 1 = 6.

2 digits: C5 takes 2× p22+C5 takes 1× p 1 1 = 25.

3 digits: C5 takes 3× p33+C5 takes 2× p22× 2 = 100.

4 digits: C5 takes 4× p44+C5 takes 3× p33× 3 = 300.

5 digits: C5 takes 5× p55+C5 takes 4× p44× 4 = 600.

6 digits: 5× p55 = 5× 120 = 600.

The total is 163 1.

Here, the calculation method will be explained first, such as two digits: C5 takes 2× p22+C5 takes 1× P 1 1 = 25.

First, take two permutations from five numbers that are not zero, that is, C5 takes 2×P22. In another case, one of the five non-zero numbers matches 0 to form a two-digit number, that is, C5 takes 1×P 1 1. Because 0 can't be the highest bit, there are only 1 possibilities for the highest bit.

(3) How many odd numbers of six digits can be formed? (288)

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The analytic higher order cannot be zero, and the odd digits are 1. If 3,5, consider the low order first and then the high order, that is, 3× 4× p44 = 12× 24 = 288.

(4) How many four digits are divisible by 25? (2 1)

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There are two possibilities to analyze four digits divisible by 25.

The last two digits are 25: 3× 3 = 9.

The last two digits are 50: p42 = 4× 3 = 12.

* * * 9+ 12 = 2 1.

(5) How many numbers greater than 20 1345 can be formed? (479)

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analyse

Judging from the six digits of the number 20 1345, it is the smallest six digits and the highest digit is 2, so what are the six digits with the highest digit greater than or equal to 2?

4× p55 = 4× 120 = 480 If the number 20 1345 is removed, 480- 1 = 479 is greater than 20 1345.

(6) Find the sum of all the three digits. (32640)

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Analyze every position to analyze it.

Sum for the hundredth time: m1=100× p52 (5+4+3+2+1)

Sum of the 10th digit: m2 = 4× 4×10 (5+4+3+2+1)

Unit sum: M3=4×4(5+4+3+2+ 1)

Sum m = m 1+m2+m3 = 32640.

5. A product produced 100 pieces, of which 2 pieces were defective, and now 5 pieces are selected for inspection.

(1) How many drawing methods are there for "just two defective products"? ( 152096)

Analysis means that 3 out of 5 samples are qualified, that is, they are taken out from 98 qualified samples.

So C2 is 2× c98,3 =152096.

(2) How many methods are there to draw only one defective product? (7224560)

The analysis is the same as above. First, select 1 defective product from 2 defective products, and then select 4 qualified products from 98.

C2 takes1× c98,4 = 7224560.

(3) How many drawings are there for "no defective products"? (679 10864)

The analysis is to extract five C98s from 98 qualified ones and take 5 = 679 10864.

(4) How many ways can you draw "at least one defective product"? (7376656)

There are at least 1 permutations by analyzing all permutations and then removing the permutations without defective products.

C 100 takes 5-C98 takes 5 = 7376656.

(5) How many drawing methods are there for "at most one defective product"? (75 135424)

Analyze all the arrangements, and get rid of two defective products, that is, at most one defective product.

C 100 takes 5-C98 takes 3 = 75 135424.

6. Randomly select 3 sets from 4 sets of A and 5 sets of B TVs, in which A and B TVs are at least 1 set, and the different method is * * * ().

(A) 140 species (B)84 species (C)70 species (D)35 species.

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According to the conditions, we can be divided into two situations.

The first case: 2 sets of A+ 1 set of b, that is, C4 takes 2×C5, 1 = 6× 5 = 30.

The second case: 1 set A+2 sets B, that is, C4 takes 1×C5 takes 2 = 4× 10 = 40.

So the total is 30+40 = 70.

7.50 pieces of products, 4 pieces are defective, there are _ _ ways to extract 5 pieces from them, and at least 3 pieces are defective.

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If there are at least 3 pieces, it means 3 or 4 pieces.

3 pieces: C4 is 3×C46, and 2 = 4 140.

4 blocks: C4 is 4×C46, 1 = 46.

* * * The electricity meter is 4 140+46 = 4 186.

8. For the three tasks of Party A, Party B and Party C, there are two people in Party A, and each of Party B and Party C 1 person. Four people are selected from 65,438+00 to undertake these three tasks, and the different selection method is (c).

(A) 1260 species (B)2025 species (C)2520 species (D)5040 species.

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The analysis is gradually completed

Step 1: firstly, select 4 people from 10: select 4 people from C 10 = 2 10.

Step 2: The work assigned to Party A and Party B is that C4 takes 2×C2, 1×C 1,1= 6× 2×1=12 boxes.

Then according to the step-by-step principle, the multiplication relation is 2 10× 12 = 2520.

9. 12 students went to three different intersections to investigate the traffic flow. If there are 4 students at each intersection, the different allocation scheme is _ _.

C(4， 12)C(4，8)C(4，4)

_ _ _ species

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Analyze each intersection and consider it in order.

Take the first crossing C 12, take 4.

The second intersection is C8-4.

The third intersection is C4 4.

The result is that C 12 takes 4×C8, and 4×C4 takes 4.

Maybe someone here will say that three different roads don't need P33. Actually, it's not. When we randomly select the number of people from 12, in fact, these classification situations already include different roads. If we× P33 again, we will think twice.

If we don't consider the differences of intersections here, that is, they are all the same intersection, the situation will be different, because we consider the differences of intersections when allocating the number of people. So we have to get rid of this possibility finally, so in the case of the above results, it should be ÷P33.

10. There are eight programs in a program table. How many arrangements can be made by adding three more programs if the relative order of the original programs remains unchanged? 990

analyse

This is a method of permutation and combination, called quadratic interpolation.

It is troublesome to answer directly. You can insert 9 blanks with one program first, and there are P(9, 1) methods. There are p (10, 1) ways to insert10 with another program; Using the last program to insert 1 1 vacancies, there is a method of P( 1 1, 1), and using the multiplication principle, it is obtained that all different addition methods are p (91) × p (/kloc).

In addition, there are three new programs at the position of 1 1 3, and then the original eight programs are supplemented at the remaining eight positions, and there is only one solution, so all methods have P 3 1 1 × 1.