First, the number of elements in the set is the most common full arrangement. If you use 1, 2, 3, 4, 5, 6, 7, 8, 9 to form 9 non-repeating numbers, then every 9 numbers are an element of set A, and there are 9 * * in set A! Element. In the following, we use S(A) to represent the number of elements in the set.
Second, the corresponding relationship of sets There is a corresponding relationship between the two sets (the concept of the function that I learned before is the corresponding relationship of sets). If set A corresponds to set B one-to-one, then S(A)=S(B) If each element in set A corresponds to n elements in set B, the number of elements in set B is n times that of a (strictly defined as dividing set B into several subsets, each subset has no * * * identical elements, and the number of elements in each subset is n, then the subset becomes an element of set B, and the elements of A correspond to the subset of B one-to-one).
Example 1: 1, 2, 3, 4, 5, 6, 7, 8, 9, the six-digit set A is a nine-digit set which is not a complex number, and S(A)=9! Set b is a set of six digits, and their numbers are not repeated. Set A is divided into subsets, and the rule is that the elements with the same first 6 bits form a subset. Obviously, there are no * * * elements in each subset. The number of elements in each subset is equal to the whole arrangement of the remaining 3 numbers, that is, 3! At this time, the elements of set B and the subset of A are in one-to-one correspondence, so S(A)=S(B)*3! S(B)=9！ /3! This is the P (9,6) that we found by the previous method.
Example 2: Choose 6 players from the players numbered 1-9 to form a team. How many ways are there? Let the set formed by different selection methods be C and the set B be a six-digit set with non-repeated numbers. Set b is divided into subsets, and the rule is that all numbers composed of the same number form a subset, then each subset is a complete arrangement of some 6 numbers, that is, each subset has 6! Element. At this time, the elements of the set C and the subset of B are in one-to-one correspondence, so S(B)=S(C)*6! S(C)=9！ /3! /6! This is what we found by the previous method. C (9, 6) and above are simple examples, so it doesn't seem necessary to be so complicated. But the concept of set is the source of permutation and combination formula, and it is also a deeper understanding of the formula. You may not realize that when we usually count, we say 1, 2, 3, 4, 5, a * * *, a total of five items. At this point, we are establishing a one-to-one correspondence between the set of items and the set (1, 2, 3, 4, 5), precisely because the number of items and the set (65438. The purpose of writing this article is to clarify these potential ideas so that I can use them to solve more complicated problems.
Example 3: Nine people sit in a circle and ask how many different sitting postures there are. 9 people in a row, there are 9 different ways! The difference between a circle of nine people is that there is no starting point and no ending point. Let set D be a set of sitting methods sitting in a circle. Starting from anyone, expand the circle into a straight line, corresponding to different elements in set A, but equivalent to the same sitting method in set D, so each element in set D corresponds to 9 elements in set A, so S(D)=9! /9 The method I said in another post is to fix one person first, and then rank others. The result is 8! . This method actually finds the corresponding relationship between set A and set D. The key to solving the problem with the idea of set is to find the corresponding relationship between sets, so that a subset of one set forms a one-to-one correspondence with the elements of another set.
Example 4: Use 1, 2, 3, 4, 5, 6, 7, 8, 9 to form a 9-digit non-repeating number, but it is required that 1 be placed before 2, so as to find the number of 9-digit numbers that meet the requirements. Set a is a complete arrangement of 9 numbers. Set A is divided into set B and set C. In set B, 1 comes before 2, and in set C, 1 comes after 2. Then S(B)+S(C)=S(A) establishes the following correspondence between set b and set c: the number formed by exchanging the positions of any element 1 and 2 in set b corresponds to the same number in set c, then this correspondence is one-to-one. So S(B)=S(C)=9! /2 In the same way, we can solve the following problems: Choose three different numbers from the nine numbers 1, 2, 3…, 9 as the coefficients of the function y=ax*x+bx+c, which requires a >；; B>c, how many such functions are there?
Example 5:M balls are packed in N boxes in different packaging ways, and the boxes are arranged in sequence. We have discussed this issue, so I'll put it in a more vivid way. Suppose we connect M balls in a row with thin thread, and then cut off the thin thread with N- 1 knife, then we can divide M balls into N groups in sequence. Then each loading method of m balls into n boxes corresponds to a wire cutting method. The line cutting method is equal to the arrangement of m balls and N- 1 knives (if two knives are arranged together, the number of balls in the corresponding box is 0). So the total number of methods is C(M+N- 1, N- 1). Example 6: Seven people sit in a row to take pictures. The order of A, B and C cannot be changed and are not adjacent. How many arrangements are there?
Solution: Party A, Party B and Party C divide the other four people into four parts. Let the number of people in the four parts be X 1, X2, X3 and X4 respectively, where X 1, X4 "= 0, X2 and X3" 0. If the other four people are regarded as the same first, then the different arrangement is the equation X 1+X2+X3+. X3=Y3+ 1 is the number of nonnegative integer solutions of X 1+Y2+Y3+X4=2, which corresponds to the method of putting two balls into four boxes, and the number is C (5 5,3) =10. Because the other four people are different, each arrangement above corresponds to the full arrangement of four people. , so different arrangements are c (5 5,3) * 4! =240 species. After skillfully using the set method, you don't need to set the set in detail every time, but you should have a clear correspondence in your mind.