Classification and step-by-step are complementary. When you are doing a question, you usually consider classification first and then consider step-by-step.
Example 1: The 5 programs originally scheduled for a new year ’s party have been arranged into a program list, and 2 new programs were added before the start of the performance. If these 2 new programs are inserted into the original program list, the number of different insertion methods Is ()
A.42 B.30 C.20 D.12
[Answer] A. Zhonggong analysis: insert the newly added 2 programs into the original 5 programs, there are 6 kinds of interpolation methods for the first one, and there are 7 empty when inserting the second program, a total of 7 kinds of interpolation methods The problem cannot be solved, so multiplication is used in stepwise calculation, so different interpolation methods total 6 × 7 = 42 (species).
Example 2: Now three people are selected from the four persons A, B, C, and D to operate the three machines A, B, and C. It is known that A cannot operate machine A, and B can only operate machine C. Both C and D can skillfully operate these three machines. Ask how many arrangements there are.
A.5 B.6 C.7 D.8
[Answer] D. Zhonggong analysis: We can think about this question in this way: C machine must be operated by someone. If I choose B, he can only operate C machine. If I do not choose B, C machine will arrange for others to operate . So it can be divided into two categories:
(1) There are B among the three people, and the remaining two are not sure at this time, but because the machine must be operated by someone, think from the perspective of the machine. First, the machine B is operated by B, and there is only one method; then A The machine cannot be operated by A, so choose one person from C and D to operate machine A. There are two methods. The remaining B machine can choose any one from the remaining two people. There are also two methods. Multiplying in steps, the number of methods is 2 × 2 = 4 kinds.
(2) There is no B among the three people, that is, three people are selected, and A can not be operated by A. At this time, only one person can be selected from Bingding. There are 2 methods. The B machine is random, and one of the remaining two people is selected. There are 2 methods, the last one to operate the C machine. There are 2 × 2 = 4 methods for multiplication step by step.
According to the classification, the total number of methods is 4 + 4 = 8. Choose D.
From the point of view of the solution of these two examples, the difference between classification and step is the key to solving this kind of permutation and combination problem. The classification needs to be based on the elements with restrictions. When you practice, you should think more about the restrictions in the permutation and combination problems.