Permutations and combinations are common test questions in the relationship between the number of tests, most students think that as long as the difficulty of the questions involving permutations and combinations is generally high, they are reluctant to touch, but there are some questions that we can master if we can master the common methods, we can make the seemingly complex questions easy to solve, today I will introduce you to a common method of permutations and combinations - the interpolation method.
First, the basics
1.Application environment: The interpolation method is generally used to solve the problem that there are elements in the permutation and combination that are not adjacent to each other.
2.Solution: (1) Arrange the position of other elements other than the non-adjacent elements (there are no non-adjacent elements);(2) After that, the non-adjacent elements are interpolated.
2. Demonstration of sample questions
Example] Five students: A, B, C, D, and E are arranged in a row to perform the program, if A and E are not adjacent, how many different arrangements are there?
a.48 b.72 c.96 d.120
Idea pointing] Stem** shows that "A and E are not adjacent", combined with the application environment of the interpolation method, it can be considered when there are elements that require not to be adjacent. First, the other elements other than the non-adjacent elements, i.e., B, C, and D, are arranged firstAfter that, the three elements B, C, and D can be formed into a total of 4 voids, and the non-adjacent A and E can be inserted into any 2 of the 4 airs, so as to ensure that A and E must not be adjacent.
Analysis] Answer: B.
According to the idea of solving the example problem, let's do a simple exercise with another problem.
Practice] There are 5 boys and 5 girls in a certain interest group, and they all prepare performances. Four students are now selected to perform one program each, and there must be both boys and girls among the four students, and there can be no consecutive performances by boys. So how many different programming arrangements are there?
a.3600 b.3000 c.2400 d.1200
Analysis] Answer: c.
In the question, "the program cannot be performed continuously by boys", that is, the boys are not next to each other, and the interpolation method can be considered. Combined with the requirements of the question stem, 4 students need to be selected to perform, and there must be both boys and girls among the 4 people. In the first case, if there is 1 boy and 3 girls, there is no case where the boys are next to each other.
Through the demonstration and practice of the above example questions, I hope that everyone will be less afraid of the permutations and combinations of questions, and try more, and practice more on the basis of understanding and mastering the common methods, and will definitely gain something. Finally, I wish you all the best in your exams!