Problem-solving skills of permutation and combination 12 method First of all, talk about the general problem-solving rules of permutation and combination comprehensive problems: 1) Whether to use "classification counting principle" or "step-by-step counting principle" depends on the way we complete a thing. When we can classify and complete it, we will use "classification counting principle" when we need to complete it step by step; So, how to determine whether it is classified or step by step? "Classification" means that any one of them can complete a given event independently, while "step by step" needs to complete all the steps of a given event. Therefore, to accurately understand the two principles, it is emphasized that several methods to accomplish a thing do not interfere with each other, are independent of each other, cross each other into an empty set, and integrate into a complete set. No matter which method is used, things can be done independently. The principle of step-by-step counting emphasizes that all steps are indispensable, and all steps need to be completed in order to complete this thing. 2) The definitions of permutation and combination are similar, and their difference lies in whether they are related to order. 3) Complex permutation problems are often visualized by experiments, and "tree diagram" and "block diagram" are drawn to seek solutions. Because the correctness of the results is difficult to test, it is often necessary to solve them in different ways to get the test. 4) Classification according to the nature of elements and gradual progress according to the continuity of events are the basic thinking methods to deal with the problem of permutation and combination, and attention should be paid to the meanings of restrictive words such as "at least, at most". 5) The general idea is to select elements (combinations) first, then arrange them, "classify" according to the nature of elements and "step by step" according to the process of events, which is always the basic principle and method to deal with the problem of permutation and combination. Through problem-solving training, pay attention to accumulate and master the basic skills of classification and gradual progress, ensure that each step is independent, and make the classification standard clear, and the level of gradual progress is clear, with no emphasis or omission. 6) When solving the synthesis problem of permutation and combination, we must deeply understand the concept of permutation and combination, be able to classify the problems skillfully, and keep in mind the formula of permutation and combination number and the nature of combination number. The common mistakes are duplication and missing counting. In short, the basic laws to solve the permutation and combination problem are: classification addition, step-by-step multiplication, clear permutation and grouping, and clear addition and multiplication; Orderly arrangement, disorderly combination; If it is difficult, it will be reversed and indirectly ruled out. Secondly, while grasping the essential characteristics and laws of problems, we should flexibly use basic principles and formulas to analyze and solve them, and at the same time pay attention to some problem-solving strategies and methods, so that some seemingly complicated problems can be solved easily. Here are several commonly used problem-solving methods and strategies. First, the "priority arrangement method" of special elements (bits): for the arrangement and combination of special elements (bits), the special is generally considered first, and then others are considered. Example 1. Use five numbers, 0, 2, 3, 4 and 5, to form a three-digit number. There are no duplicate numbers, and even numbers * * * have (). A. 24 B.30 C.40 D.60 [Analysis] Because three digits are even, the number at the end must be even, and because 0 can't rank first, 0 is one of the "special" elements and should be given priority. It is divided into two categories: the last 0 and the last 0: 1). At the end of line 0, there is an A42. Then there is C2 1 A3 1A3 1. According to the principle of fractional counting, * * has even numbers A42+c2 1a3 1 = 30, so B. 2 is selected. Total elimination method: for negative problems, unqualified ones can be eliminated from the total. For example, in the case of 1, this method can also be used to solve the problem: there is an A53 in the full arrangement of five numbers forming three digits. After arrangement, it is found that 0 can't be ranked first, and numbers 3 and 5 can't be ranked last. These two permutations should be excluded, so there are A53-3A42+C21A31= 30 even numbers. 3. Reasonable classification and accurate step-by-step classification of the constrained permutation and combination problems, and step-by-step classification according to the nature of the elements and the continuous process of things, so that the classification standards are clear, the step-by-step levels are clear, and no weight is missed. 4. Binding method of adjacent problems: When solving the problem of requiring some elements to be adjacent, first consider the whole, "bind" the adjacent elements into a "big" element and arrange them with other elements, and then consider the order between elements in the big element. The binding method is the solution strategy. Among them, there are 3 math books, 2 foreign language books and 3 books on other subjects. If these books are arranged in a row on the bookshelf and math books are put together, there are () ways to put foreign documents together. Solution: bind three math books together as a big book and two foreign language books together as a big book. The other three math books are arranged in A33, and two foreign language books are arranged in A22. According to the principle of step counting, there is an arrangement method A55 A33 A22= 1440 (species). Note: When using the binding method to solve the permutation and combination problem, we must pay attention to the internal order of "binding" large elements. 5. "Interpolation" is used for nonadjacent problems: nonadjacent problems mean that some elements cannot be adjacent, and they are separated by other elements. In order to solve this kind of problem, we can insert the specified non-adjacent elements into the gaps and positions at both ends, so it is called interpolation. Example 3, use 1, 2, 3, 4, 5, 6, 7, 8 to form an eight-digit number, which is not plural. It is required that 1 be adjacent to 2, 2 to 4, 5 to 6 and 7 to 8. There are eight digits like () * * *. Solution: Since 1 is required to be adjacent to 2 and 2 to 4, the three numbers 1, 2 and 4 can be tied together to form a large element, and only 2 can be arranged in the middle of this large element, and 1, 4 can be arranged on both sides, so there are also A22 arrangement methods in the element *. Arrange these three elements first. There are 33 ways to arrange them. Then choose two from the gaps formed by the three elements arranged in front and the four positions at both ends, and insert the numbers 7 and 8 which are not adjacent. * * * There are A42 interpolation methods. So the qualified eight-digit * * * is A22A33A42 = 288 (species). Note: When solving nonadjacent problems by interpolation, pay attention to whether the position to be inserted includes the positions of both ends. 6. Ordering by "division": For the problem that some elements are arranged in a certain order, we can first arrange them all together with other elements, and then divide them by the total arrangement number. Example 4: Six people line up, and A, B and C line up in the order of "A-B-C". How many kinds of queuing methods are there? Analysis: There are A66 queuing modes regardless of additional conditions, and only one of A33 queuing modes of A, B and C meets the requirements. So there are A66 ÷A33 = 120 eligible permutation methods. (or A63) Example 5. Four boys and three girls are not equal in height. Now line them up. Please arrange the girls from left to right from short to high. How many ways are there? Solution: First, four of the seven positions were given to boys, with the arrangement of A74, and the remaining three positions were given to girls, with only one arrangement, so there was the arrangement of A74. (It can also be A77 ÷A33) Seven. The problem of arrangement is "in-line arrangement": the problem of arranging several elements into several rows can be handled by a unified arrangement method. Example 6. Seven people sit in two rows, three in the first row and four in the second row. How many different sitting positions are there? Analysis: Seven people can sit in the first two rows at will, and there are no other conditions, so the two rows can be regarded as one row, and there are 77 different sitting methods. Eight. Item-by-item test method: When the additional conditions in the questions increase and the difficulties are directly solved, the rules are gradually found through experiments. Example 7. Fill the number 1 in the box with the number 1, 2, 3, 4, and 1 in each box. The number of filling methods for different numbers in the box is () a.6b.9c. 1 1. If the second box is filled with 3 or 4, then there is only one filling method for the last two boxes. There are nine ways to fill a * * *, choose B 9. Building a model "partition method": For complicated layout problems, we can design another scene and build a partition model to solve the problem. Example 8. How many positive integer solutions does the equation a+b+c+d= 12 have? Analysis: Establish baffle model: arrange 12 identical balls in a row, insert 3 baffles randomly in the gap between them, and divide the balls into 4 piles. The number of balls in each pile obtained by each division corresponds to a set of positive integer solutions of A, B, C and D, so the number of groups of positive integer solutions of the original equation. X. Exclusion method: For the permutation and combination question with "at most" or "at least", if the direct answer needs complicated discussion, you can consider "overall impurity removal", that is, the method of deleting unqualified permutations or combinations in the group, so as to calculate the number of qualified permutations and combinations. Example 9. Randomly take out 3 TV sets from 4 A and 5 B TV sets, including at least one A and B TV set. Then there are () different solutions. A. 140 kinds of B.80 kinds of C.70 kinds of D.35 kinds of solutions: among the three extracted tables, the extraction method that does not contain A or is not suitable for B is not satisfactory, so the extraction method that meets the question is C93-C43-C53=70 (kinds). So C. note: this method is suitable for exercises with clear negative situation and simple calculation. XI。 Step-by-step exploration method: For the problems with complex situations and difficult to find their laws, we need to carefully analyze them, and take out the natural numbers with the law of 1 00 from1to 100 at one time, so that the sum is greater than 100. Solution: In the addition of two numbers, the smaller number is the addend,1+100 >; When 100 and 1 are addends, 1, 2 is addend, …, 49 is addend, 50 is addend, but 5 1 is addend, 48 is addend, …, 99. Therefore, there is a corresponding method of (1+2+3+…+50)+(49+48+…+1) = 2500 species 12 pairs1:for example,1. 100 Single cycle elimination among players. Solution: To produce a champion, all players except the champion must be eliminated, that is, 99 players must be eliminated and 1 player must be eliminated, so there are 99 games.