"basic principles"
Addition principle: There are n different ways to accomplish a thing, and each way has many possible ways. Then, to complete this, you need to add up these possible practices; Multiplication principle: it takes n steps to complete one thing, and each step has m 1, m2, …, mn. Then we need different methods:: m 1×m2×…×mn to accomplish this.
"permutation and combination"
Arrangement: from N different elements, any M () elements (the selected elements here are different) are arranged in a column in a certain order, which is called the arrangement of extracting M elements from N different elements.
Combination: M () elements of N different element species are combined into a group, which is called the combination of M elements of N different elements.
"The difference between arrangement and combination"
Combination is to select m elements from n different element species, and how many different selection methods there are. Only m elements are selected, regardless of the order of these selected elements; Arrangement should not only select, but also arrange the selected elements in order, that is, consider the order of the selected elements. So from this point of view, the number of combinations must not be greater than the number of arrangements.
"Special problem solving methods"
There are several special methods to solve the problem of permutation and combination: interpolation method and plug-in method. The following is explained one by one:
(1) interpolation method
This kind of problem generally has the following characteristics: there are elements with constant relative position in the topic, which may be called fixed elements, and there are elements with changing relative positions, which are called active elements, and what we are required to do is to insert these active elements into the air formed by fixed elements. For example:
Example 1: There are three original programs in a program list. If the relative order of these three programs remains unchanged, two new programs will be added. How many arrangements are there?
12 C.6 D.4
Solution 1: There are three "fixed elements" and two "active elements", but pay attention to the order of the active elements themselves. When two new programs are adjacent, there are four ways to put them in the air formed by "fixed elements": C4 1x2 = 8. 2)。 When two programs are not adjacent, it becomes an arrangement problem, that is, two programs are randomly selected from four airspace and two different programs are played in turn. There are four methods: P42= 12. To sum up, 12+8=20 kinds of * *.
Solution 2: partial solution. 1) You can insert a program first in four ways; 2) Then insert another program, and the program inserted for the first time will also become a "fixed element", so * * * has five spaces to choose from; Apply the principle of multiplication: 4×5=20 kinds.
Example 2. Xiaoming lives on the second floor. Every time he goes home and goes up the stairs, he takes two or three steps at a time. Given that there are 16 steps between adjacent floors, how many different ways does Xiao Ming walk from the first floor to the second floor?
A.54 B.64 C.57 D.37
Solution 1: list solution, the fourth number = the first number+the second number.
Step123456789101213141516.
Take 011122345791216212837.
Solution 2: interpolation method to solve the problem: consider the number of steps taken:
1) has 0 steps (that is, all 2 steps), so there are 1 kinds of walking methods;
2) There are 1 steps. (impossible to complete the task);
3) three steps twice, that is, two steps five times:
(a) When three steps are adjacent twice: it is equivalent to putting two adjacent three steps into the air formed by five two steps, and there are six ways (C 61= 6);
(2) When three steps are not adjacent twice: it is equivalent to putting two non-adjacent three steps into the air formed by five two steps, and there are C62= 15 ways.
4) Three times (impossible)
5) If you take three steps and four times, you will take two steps and two times, and the roles will be exchanged, thinking about putting two steps in the air formed by three steps. As in (3), there are five ways to go, C51+C52 =1,considering two cases of being next to or not.
6) There are 5 times (impossible), so there are * * *:1+6+15+15 = 37 kinds.
(2). Plug-in method: generally solve the problem of the same element distribution, and the restrictions on the divided elements are very weak (generally not equal to zero), and only the number of shares is required.
Example: example 1. Distribute 20 computers to 18 village, requiring each village to distribute at least one computer. How many distribution methods are there? Analysis: The idea of this question is the idea of patch panels: randomly insert 17 boards into 20 computers to form 19 air, and divide it into 18 copies. Then * * has:
C1917 = c192 =171species. Eg2。 There are 10 tablets, and you should eat at least 1 tablet every day until you finish eating. * * * How many different ways can there be?
Solution 1: 1 day: there are C90= 1 species; Finished in 2 days: there are 9 kinds of C 91=;
10 days later: C99= 1 species; Therefore * * * has: c90+c91+…+c99 = (1+1) 9 = 512.
Scheme 2: 10 computer has 9 vacancies, and each hole can be inserted or not, that is, each hole has two choices, * * * has 9 vacancies, and * * * has 29=5 12. Only two special methods in permutation and combination are discussed here. As for other questions, please refer to other Chinese books. Com, so I won't go into details here.
"Application of permutation and combination in other problems"
Examples. The school has prepared a 1 152 square color board to make a rectangle. How many different spellings are there?
A.52 B.36 C.28 D. 12
Solution 1: This problem actually wants to decompose 1 152 into the product of two numbers, so1152 =1×152 = 2× 576 = 3.
Solution 2: (Solve with permutation and combination knowledge)
From 1 152=27×32, then what we need to do now is to divide these 7 2s and 2 3s into two parts. When they are evenly distributed, the length and width of the rectangle will be fixed.
Specifically: 1) When two 3s are together, there are eight distribution methods (from 0 2s to 7 2s); 2) When two 3s are not together, there are four distribution methods, that is, one 3s is followed by 0, 1, 2 and 3 2s. Therefore, * * * has 8+4= 12.
Solution 3: If 1 152=27×32, all products of 1 152 are 1 152, and the number of factors is (7+1)×.