The analysis of the above problem-solving ideas can be summarized as follows: (1) Examining questions, grouping and distinguishing; Reasonable classification, using quasi-addition and multiplication; Considerate, leak-proof and heavy-proof; Direct and indirect, ideas can be followed; Element position, special priority; Answer more questions to test the authenticity.
The following is a strategic analysis of several typical permutation and combination problems, in order to find an effective way to solve the corresponding problems.
First, special priority, usually in the back.
Priority should be given to the special elements and special positions in the problem. In operation, for practical problems, sometimes it is "element priority" and sometimes it is "position priority".
Example 1 0, 2, 3, 4, 5, these five numbers make up three numbers without duplicates, among which there are even numbers * * *?
Scheme 1: (element priority) is divided into two categories: the first category, with A42 0 in the unit and A2 1 A 3 1 0 in the tenth place; The second category, excluding 0, includes A2 1 A32.
Therefore, * * has (a42+a21a31)+a32a21= 30.
Note: When considering each category, one location should be given priority.
Scheme 2: (location priority) is divided into two categories: the first category, with A42 types of zeros in the unit; In the second category, 0 is not a single digit. First, choose one of the two even numbers to put one digit, then choose one to put one hundred digits, and finally consider ten digits. There are A2 1A3 1A3 1.
Therefore, * * has a42+a21a31a31= 30.
The exercise 1( 1989 nationwide) consists of the numbers 1, 2, 3, 4 and 5. There are five numbers without duplicate numbers, and even numbers * * * are less than 50000 (answered by numbers).
Answer: 36
Second, the row is mixed, and the back row is selected first.
For the mixing problem of permutation and combination, it is advisable to combine the selection elements first and then arrange them.
Example 2( 1995 nationwide) Put four different balls into four boxes numbered 1, 2, 3 and 4. How many ways can I put an empty box?
Solution: From the meaning of the question, there must be two balls in a box. The balls in the same box are combined together, and different balls are arranged in different boxes. So there is a broadcast mode of C42A43= 144.
Exercise 2 consists of the numbers 1, 2, 3, 4, 5, 6 and 7. There are three odd numbers and two even numbers. How many numbers are not repeated?
Answer: yes, C43C32A55= 1440 (pieces)
Third, the elements are adjacent and treated as a whole.
For the problem that some elements need to be arranged adjacently, we can first bind the adjacent elements into a whole and treat them as one element, and then arrange them with other elements, and at the same time arrange the adjacent elements themselves.
Example 3 Five boys and three girls lined up and asked the girls to line up together. How many arrangements are there?
Solution: First tie three girls into a whole, and then line up with five other boys. At the same time, the three girls themselves should be fully arranged. There are A66 A33 multiplication principles.
Exercise 3 Four pairs of brothers and sisters stand in a row. How many ways can each brother and sister stand together?
Answer: A44 24 = 384
Four, element spacing, bit insertion
For some elements that need interval arrangement, insert method is used.
Example 4 Five boys and three girls are arranged in a row, requiring that the girls are not adjacent to each other and cannot be arranged at both ends. How many arrangements are there?
Solution: Boys with unlimited conditions rank first, and girls are inserted in four gaps between five boys. According to the principle of multiplication, there are A55A43 kinds.
Note: ① The question of "who inserts who" must be distinguished. Arrange the elements with unlimited conditions first, and then insert the elements that must be spaced; ② Count the number of positions that can be inserted; ③ When inserting, it is necessary to grasp whether it is combined insertion or arrangement insertion.
Exercise 4 Four men and four women stand in a row. How many ways do men and women stand alternately?
Answer: 2A44 A44
There are 9 street lamps numbered 1, 2, 3, …, 9 on the road. Now, if you want to turn off three of them, but you can't turn off two or three adjacent lights at the same time, and you can't turn off the street lights at both ends, how many ways can you turn off the lights that meet the requirements?
Solution: Because there are 6 strong lights and 3 dark lights in the problem, both ends can't be dark, so it is enough to insert 3 dark lights in 5 gaps of 6 strong lights, and there are C53 kinds.
Exercise 5 Choose three non-adjacent natural numbers from the ten numbers 1, 2, …, 10. How many different ways are there?
Answer: C83.
5. Ordering of elements, that is, arranging first and then dividing, or choosing the position not to arrange, or arranging first and then inserting.
For the arrangement of some elements with fixed order, we can arrange them all first, and then divide them by the overall arrangement of sorted elements, or choose the position of sorted elements from the total positions that do not participate in the arrangement, and then arrange other elements. You can also put ordered elements first and then insert other elements one by one.
Example 6 5 people participated in the 100 meter race. If they don't arrive at the finish line at the same time, what are the situations when A arrives earlier than B?
Scheme 1: First, there are A55 species for 5 people in the whole platoon. Because there are A22 species in the whole row of A and B, and only 1 species here meets the requirements, we have to divide them by A22 species in the whole row of sequencing elements, so there are A55/A22=60 species.
Scheme 2: First, choose two of the five locations to put C52 sequencing elements (A and B), and then arrange for the other three people to have A33. According to the principle of multiplication, * * has C52A33=60 kinds.
Scheme 3: Fix A and B first, then insert the first person in the other three, then insert the second person in four ways, and finally insert the third person in five ways. From the principle of multiplication, there are 3×4×5=60 kinds of * *.
Exercise 6: Make up a performance program, and arrange six dance programs successively. How many ways are there to insert five singing programs?
Answer: a11/a66 or c116a55 = c115a55 or 7× 8× 9×10.
Six, "small groups" arrangement, "groups" before the whole.
When some elements in some arrangement problems need to form a "group", they can be "grouped" according to constraints, regarded as one element, and then arranged with other elements.
Four male singers and two female singers jointly held a concert. The order of performance requires two male singers between two female singers. How many appearance schemes are there?
Solution: First, choose two from four male singers and arrange them among two female singers. There are 22 species of A42A, and there are 33 species that regard this small group of "women and men" as 1 person, and then arrange with two other men to form A33A. According to the principle of multiplication, there are 33 kinds of A42A * *.
Exercise 7 Six people stand in a row. How many ways can a child stand between his parents?
Answer: A22 A44
Seven, different elements into the box, first pile back.
For different elements, put them in several different boxes. When there are not less than 2 elements in some boxes, you can't enter them in batches. You must pile them up before discharging them.
Five teachers are divided into three classes to engage in activities, with at least one in each class. How many different ways are there?
Solution: First, divide the five teachers into three groups, namely C5 1, 1, 2, 2, and then arrange them into three classes, namely A33, so * * has (C53+C.
Note: Different teachers are not allowed to enter the same class in batches, and must be in place at one time (otherwise there will be double counting). That is, "elements in the same box must enter at one time".
There are six students in Exercise 8. Find out the number of allocation methods in the following cases:
(1) math group assigned 3 people, physics group assigned 2 people, chemistry group assigned 1 person;
(2) The math group is assigned 2 people, the physics group is assigned 2 people and the chemistry group is assigned 2 people;
③ Divided into three groups: mathematics, physics and chemistry, with 3 people in one group, 2 people in the other group and 1 person in the other group;
④ Divide into three groups for volleyball training.
Answer: ① c63c32c11; ②c62c 42 c 22； ③c 63 c 32 c 1 1 A33； ④C62C42C22/A33 .
Eight, the same component into the box, separated by baffle.
Example 9 10 tickets for visiting the park are distributed to five classes, and each class has at least 1 tickets. How many choices are there?
Solution: this is just the number of votes, not the order. Therefore, 10 tickets can be regarded as 10 identical balls, which are put into five different boxes, and each box has at least 1 balls. You can arrange 65,438+00 balls in a row, and then insert four "baffles" in four of the nine intervals to divide them into five grids (forming five grids).
Note: The baffle separation model is specially used to solve the distribution problem of the same element.
Exercise 9 Choose 12 people from 10 classes in the whole school to form a volleyball team, with at least one person in each class. How many ways are there?
Answer: C 1 19
Nine, the arrangement of the two types of elements, using the method of combining location selection.
Example 10 10 stairs need 7 steps, and each step can span one or two steps. How many different generation methods are there?
Solution: according to the meaning of the question, there are 4 steps across a single level and 3 steps across two levels. You can just choose 3 steps across two levels. So there are 73 ways to cross.
Note: the arrangement of the two types of elements involves a wide range, so we should pay attention to it.
Exercise 10 3 red flags and 2 yellow flags, all held on the flagpole as signals. How many different signals can you send?
Answer: C52
Example 1 1 What are the shortest routes from vertex A to vertex B along the grid lines in the graph?
Solution: For each shortest path, three | lines and four-lines should be taken, which is the unordered arrangement of the two types of elements. Therefore, there are 74 or 73 roads to go.
Example: 12 Choose 10 people from five classes to form the school basketball team (without any requirements). How many ways are there?
Solution: This problem is different from the example 12. Although it can still be regarded as four "baffles", 10 balls are divided into five squares (forming five boxes), which is a problem of disorderly arrangement of balls and baffles. But there may be no balls in some boxes, so the four "baffles" should participate in the arrangement and occupy the position like 10 balls, so there is a selection method of C 144.
What are the extensions of exercise11(a+b+c+d)10?
Tip: Since each term is a 10 power formula obtained by multiplying one or more of A, B, C and D, it can be regarded as a disorderly arrangement of 10 balls and three baffles, so * * * has the term C 133.
Note: How to equivalently transform the problem into the problem of "arrangement of two types of elements" is the key to solve the problem.
Ten, the number is not less than the number of boxes, fill in first and then divide.
Example 13 15 put the same balls into the boxes numbered 1, 2, 3, and the number of balls in the boxes shall not be less than the number of numbers. How many different ways are there?
Solution: First "fill" six balls in each box according to the quantity (meet the minimum requirements), and then put nine balls in three boxes. Two baffles can be arranged together with nine balls (that is, the arrangement of two types of elements), including C 1 12.
Eleven, multi-class element combination, classification and extraction.
There are 1 1 workers in 14 workshop, including 4 lathes and 5 locksmiths. AB can also be both lathes and locksmiths. Today, we need to mobilize four turners and four locksmiths to complete a certain task. How many different methods are there?
Solution: Different adjustment methods are divided into the following three categories according to the lathe workers: the first category is to adjust four lathe workers and four fitters; Type II: 3 turners, 4 fitters, turners from AB 1 person; In the second category, there are 2 turners and 4 fitters, and AB is regarded as a turner. Therefore, * * has different tuning methods: C44C74+C43C21C64+C42C2C54 =185.
Note: This problem can also be classified by locksmith. If we classify according to A and B, the problem will be complicated.