Recently, I was preparing for the final exam of combinatorics. When I reviewed the chapter on groups, my understanding was not very thorough. Let me record the problem-solving methods related to groups.
The definition of the group will not be introduced one by one, and the topic will be cut directly.
Table of contents
[Thoughts for solving problems with permutation and multiplication]
Two applications of Pólya's theorem
(2) There are 3 different colors of beads strung into a necklace of 4 beads, what are the options?
a permutation multiplication
Known groups P1 and P2, find P1P2 and P2P1, first take the courseware:
The blogger was puzzled when he saw this, and he suddenly realized it after consulting the information.
[Thoughts for solving problems with permutation and multiplication]
First of all, the first line of the two groups P1 and P2 can be understood as the initial sequence, while the second line is the permuted sequence.
And P1P2 is to replace P1 first, and then replace P2 on the basis of P1. P2P1 is just the opposite. Draw a picture to let us understand it more vividly.
This will make it easier to understand, let’s try P2P1~
Two applications of Pólya's theorem
Do two questions to familiarize yourself with the polya theorem
(1) The three vertices of an equilateral triangle are colored with red, blue and green. How many schemes are there?
Or look at the answers to the courseware first
【Problem solving ideas】
By reading the question, we can find that this is a case of n=3 (three vertices), m=3 (three colors) . There are 6 situations for coincident movement of equilateral triangles, namely 0°, 60° (240° in the courseware, but I feel that 60° and 240° have the same meaning), 120° and flipping through the midline of three vertices (that is, 3 cases), so there are 6 cases in total.
The following six situations are analyzed in detail:
①0°, each vertex is different and cannot be replaced with each other, so it is (v1)(v2)(v3).
②60°, the position of each vertex becomes the position of the right vertex, and the three points are exchanged, v1 becomes v3, and v3 becomes v2, so in a permutation relationship, it is (v1v3v2). [This corresponds to 240 in the courseware °The situation is (v3v2v1)]
③120°, the position of each vertex becomes the position of the left vertex, and the three points are exchanged, v1 becomes v2, and v2 becomes v3, so in a permutation relationship, it is (v1v2v3).
④ Turn over along the midline of v1, v1 does not move, and v2 and v3 are replaced, so there are two replacement relationships, namely (v1) (v2v3).
⑤Flip along the midline of v2, v2 does not move, and v1 and v3 are replaced, so there are two replacement relationships, namely (v2) (v1v3).
⑥Flip along the midline of v3, v3 does not move, and v1 and 2 are replaced, so there are 2 replacement relationships, namely (v3)(v1v2).
After the above six situations have been discussed, start to calculate the number of solutions.
The reason why the previous coefficient is one-sixth is because there are 6 cases, which is a bit weighted average. The specific calculations for each case are in square brackets. Because there are 3 colors that can be counted, the power operation is performed on the basis of 3. Because there are 3 parentheses at 0° (which can be understood as 3 replacement relationships), so it is 3 to the 3rd power. Both 60° and 120° are in a bracket, so they are two 3s. In the case of ④, ⑤, and ⑥, there are two brackets, so it is the square of three 3s.
(2) There are 3 different colors of beads strung into a necklace of 4 beads, what are the options?
Or look at the answers to the courseware first
But the blogger feels that this ppt has some narrative problems. According to the information I checked on the Internet, I will not explain it according to him. I will directly post my handwritten answer~~~
【Problem solving ideas】
Finished flowering*★,°*:.☆(￣▽￣)/$:*.°★* If you learn from it~~