Before introducing permutation and combination, let's understand the basic operation formula! C5 takes 3 = (5× 4× 3)/(3× 2× 1) C6 takes 2 = (6× 5)/ (2× 1). From these two examples, it can be seen that the formula for CM to take n is the product of n consecutive natural numbers with seed number m as the numerator and the hierarchy with n as the denominator. Through these two examples, PMN= = N natural numbers that start from m and are continuous with themselves. The essence of combination is to study "the possibility of orderly arrangement and disorderly arrangement of m(m≤n) elements starting 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. Classification is step by step with "addition" and "multiplication". Classification: "There are n ways to do one thing and finish it" is the classification of all the ways to finish 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. Step by step, according to the characteristics of the problem, determine a feasible step by step standard. Secondly, the setting of steps should meet the requirement that this matter must be completed and can only be completed after N steps are completed continuously. 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 no matter which one of them can be accomplished alone, use addition principle to find the number of ways to accomplish it. If a thing needs to be divided into n steps, these n steps are indispensable, that is, all steps need to be completed in turn, and each step has several different ways to complete. The principle of multiplication is used to find a way to achieve this. Pay attention to the following points when solving the application problems of permutation and combination: 1. The common propositional forms of constrained permutation problems are "yes" and "no", and "adjacent" and "non-adjacent". We should master the basic problem-solving ideas and methods: (1) The method of merging elements is often used to solve "adjacent" problems. This is the most commonly used method to deal with adjacency. (2) Interpolation arrangement is the most commonly used method to solve the "non-adjacency" problem. (3) The question of "being" and "not being" often involves special elements or special positions, and it is generally arranged first. (4) For the arrangement of elements with order restriction, the order restriction can be ignored first, and the result can be obtained by using the fact of the specified order after the arrangement is completed. (2) there are restrictions. Common propositional forms: "including" and "not including", "at least" and "at most" commonly used methods in solving problems are "direct method" or "indirect method". 3. When dealing with the comprehensive problem of permutation and combination, we should analyze the conditions and classify them according to the nature of the elements, so as to achieve neither weight nor leakage, and apply two principles step by step according to the occurrence process of the event, that is, solve the problem of permutation and combination. It is also the most important way of thinking. 10 exercise questions are provided for everyone to practice. 1, all three sides are integers, and the number of triangles with the maximum side length of 1 1 is (C)(A)25 (B)26 (C)36 (D)37. The difference between the two sides is less than the difference between the third side. If the largest visible edge is 1 1, then the sum of the two outer edges cannot exceed 22, because when all three edges are 1 1, the sum of the two edges is the largest. Therefore, we start with the length of one side. If it is 1 1, the length of the other side is 1658. . 1 If it is 10, the length of the other side is 10, 9, 8.2, (it cannot be 1, otherwise the sum of them is less than 1 1, and it cannot be1. 8, 7, 3 (as for the above reasons, you can see that there are rules). The total number of rules is11+9+7+.1= (1+1) × 6 ÷. -analyze the above situation. There are four options for arranging the second passenger. Knowing the last passenger is also four possibilities. According to the principle of step by step, it belongs to the multiplication relationship, that is, 4× 4× 4 = 4 3 (3) 8 different books. Choose three books and give them to three students, one for each. How many different ways are there? -analysis, step by step. After the selection is completed, the second student has only two choices, and the last student has no choice. That is, 3×2× 1, which is a step-by-step selection and 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 question is 56× 6 = 3363, and seven students line up to take pictures. (1) How many different arrangements are there in the front row or the back row of a company? (3600)-To analyze this topic, we complete the first step in two steps: the first line should be in the middle five positions. (1440)-Step 1: Determine which is B. (3 120)-The other 5 people can start from 5, and the other 5 positions meet the P principle, that is, 5× P55 = 5× 120 = 600. In the second case, A is not at the head of the line, A is in the middle position, and the other six positions satisfy P66=720. Because it is a classified discussion, the final result is the sum of the two situations, which is 2400. (1440)—— That is to say, discussNo. 1 step by step: choose position C6 as 1 = 6 No.2: If the two selected positions meet the requirements of Party A and Party B, that is, P22 = 2, the number of species meeting the requirements of Party A and Party B is 2 × 6 = 12, and the rest. According to the principle of equal left-right probability, the number of species ranked on the left is 5040 ÷ 2 = 25204, and the number 0, 1, 2, 3, 4 and 5 is used to form a number, and there is no duplication. How many four digits can (1) form? (300)-(1631)-38+0 = 253 digits: C5 takes 3× p33+C5 takes 2× p22× 2 = 1 004 digits: C5 takes 4× p44+C5 takes 3× p33× 3 = 3005 digits: C5 takes 5× p55+C5 takes 4× p44× 4 = 6006 digits: two digits: C5 takes 2×P22+C5 takes 1×P6. Firstly, two digits are arranged from five digits that are not zero, that is, C5 takes. In another case, one of five numbers that are not 0 is selected and matched with 0 to form two numbers, that is, C5 takes 1× p 16544. (288)-(2 1)-(479)-Judging from the six digits of the number 20 1345, yes. 4× p55 = 4× 120 = 480 If the number 20 1345 is removed, the number greater than 20 1345 is 480- 1 = 479 (6). Find the sum of all three digits. (32640)-sum on: m1=100× p52 (5+4+3+2+1) sum on ten digits: m2 = 4× 4×10 (5+4+3 (152096) Analysis: That is to say, three of the five pieces sampled are qualified, that is, 98 pieces are taken out of the qualified ones. Then, how many drawing methods are there, that is, C2 takes 2×C98 and 3 = 152096 (2) "There is just one defective product"? (7224560) The analysis is the same as above. First, select 1 defective product from 2 defective products, then select 4 C2 from 98 qualified products, and select 1×C98 = 7224560(3) "No defective products" from 4. How many ways are there? (679 10864) analysis, that is, how many ways are there to extract five c98s out of 98 qualified ones? Take 5 = 679 10864 (4) "at least one defective product"? (7376656) Analyze all the permutations and then remove the permutations without defective products, that is, C 100 has at least 1 species, 5-c98,5 = 7376656 (5) How many drawing methods are there for "at most one defective product"? (75 135424) There are two defective products in all the arrangements, that is, C 100 takes 5-C98 and 3 = 75 1354246 takes at most one defective product, and three TVs are randomly selected from four A-type and five B-type TVs, of which at least 1 is required. There are () (A) 140 (b), 84 (c) and C)70 (d) methods. 2×C5 species-we can divide them into two types according to conditions. = 4× 10 = 40, so the total number is 30+40 = 70. 4 out of 7.50 products are defective, and 5 of them are selected. There are _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Choose four people from 10 to undertake these three tasks. There are (C)(A) 1260 options (B)2025 options (C)2520 options (d). 5040 kinds of analysis-In this case, according to the principle of step by step, 2 10× 12 = 25209 and 12 students went to three different intersections to investigate the traffic volume. If there are four students at each intersection, the different allocation scheme is _ _ c (4, 12) c (4, 8). 4) _ _ _ _ _ _-Number of people. In fact, these classifications already include the situation of different roads. If we try again × ×P33, we will think again and again. If we don't consider the difference of intersections here, the situation is different. Because we considered the difference of intersections when allocating the number of people, we finally had to remove this possibility. Therefore, in the case of the above results, we have to present ÷P33 10/0, and there are 8 programs in a program table. If the relativity of the original program is maintained, 990 analysis shows that this is a method of permutation and combination, which is called secondary insertion method. It is troublesome to answer directly, so one program can be used to insert 9 vacancies 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). 1)=990 species. In addition, three new programs are arranged 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 3 168.