1. Basic theory:
Binding method: When encountering the problem of "adjacent elements", first bind the specified adjacent elements together to participate in the arrangement, and then loosen the binding when the relative order of the elements needs to be considered.
Common words in the question stem such as: adjacent station, connected, continuous, etc.
Insertion method: When encountering the problem of "non-adjacent elements", first sort the unrequired elements, and then insert the space in the middle of the stroke or the space at both ends.
When using the inserting method to solve the problem of permutation and combination, we must pay attention to inserting the empty position including the "middle space" and "the space at both ends" of the elements. The problem-solving process is "arrange first, then insert blanks".
It can be seen that the bundling method mainly solves the adjacent problem, while the plug-in method mainly solves the non-adjacent problem.
Second, the true analysis
Example 1 and 5 students and 2 teachers stand in a row to take pictures. If 2 teachers are required to be next to each other but not at both ends, they are different
. There are a total of arrangements:
A.1440 types B.960 types C.720 types D.480 Species
[Analysis] There are "adjacent" in the question stem, so the selected method of doing the question must be the binding method. To solve this problem clearly, it must be divided into the following steps: The first step is to sort the elements that are not required. , That is, to arrange 5 students first, there are A(5,5) methods; the second step is to "bundle" the 2 teachers together and treat them as one person and insert them into the 4 airspaces among the 5 students, namely C(4,1) methods; In the third step, the two teachers are different and need to be arranged, that is, A(2,2) methods, this thing is done. What to do step by step, according to the principle of multiplication, there are a total of A(5,5)×C(4,1)×A(2,2)=960 different arrangements. So the answer is B.
Summary: Although the binding method and the plug-in method are two different methods, they are often used in combination.
Example 2. There are 3 original programs on a program list, if the relative order of these 3 programs remains unchanged, and 2 new programs are added, how many arrangements are there?
A.20 B.12 C.6 D. 4
[Analysis] This question is a typical example question of the board method, because it is equivalent to inserting 2 new programs into the original 3 programs, so it is necessary to find out how many vacancies are.
[Analysis] The original 3 programs have been fixed, so when the original 3 programs are arranged, there is no need to mix them.
So this thing can be done step by step, you need to put the two new programs into it in the first step and the second step.
The first step is to arrange one of the programs. In the original 3 programs, there are 4 vacancies to choose from, that is, the method in C(4,1); the second step is to arrange the second program, then the first program at this time After putting it in, there are 4 programs, that is, there are 5 vacancies to choose from, so the arrangement is the C(5,1) method, and this matter is completed at this time. Step by step, so choose the original understanding of multiplication, that is, C(4,1)×C(5,1)=20 sorts, so the answer is A.
Example 3: There are 10 street lights next to a road. In order to save electricity, 3 of them are going to be turned off. Knowing that the street lights at both ends cannot be turned off, and the turned off lights cannot be adjacent, there are () different ways to turn off the lights.
A.20 B.28 C.48 D.96
[Analysis] Read the logical relationship in the question stem clearly, and properly transform the equivalent relationship before doing the question. The meaning of the question is to insert the 3 turned off lights in the middle of the 7 bright lights, and to ensure that the turned off lights are not adjacent, so this question should belong to the blanking method.
[Analysis] For the 7 lights that are on, the first and last ends of the lights cannot be turned off, so there are only 6 lights in the middle that can be turned off, that is, C(6,3)=20 kinds. So the answer is A.
Summary: The binding method and the blanking method are small skills for solving problems. They are flexible in application. They can be applied to all types of permutation and combination problems, so everyone must distinguish when to use this method.
Example 4. Two teams of couples each take a child on a 6-seater cable car. For safety reasons, the two dads can only ride on both ends. Two children can only be next to each other. How many ways are there to sit?
A.12 B.24 C.36 D.48
Two fathers A(2,2). Two children are tied up and A(3,3) with two mothers. Two children inside A (2, 2). 24 kinds in total