There were four students lining up to go upstairs, and there were several different ways to line ups

4x3x2x1, a total of 24 types, using the factorial method.

A positive integer factorial is a multiplication of 1 times 2 times 3 times 4 all the way up to the required number.

For example, if the required number is 4, then the factorial is 1 2 3 4, and the resulting product is 24, and 24 is the factorial of 4. For example, if the required number is 6, then the factorial is 1 2 3 ......6, the product obtained is 720, and 720 is the factorial of 6. For example, if the required number is n, then the factorial is 1 2 3 ......n, let the product be x, and x is the factorial of n.

Two commonly used permutations: basic counting principles and applications

1. Addition principle and categorical counting method:

Each of the methods in each class can accomplish this task independently; The specific methods in the two different types of approaches are different from each other (i.e., the classification is not duplicated); Any way to accomplish this task belongs to a certain category (i.e., classification is not missing).

2. Multiplication principle and step-by-step counting method:

None of the methods of any step can complete this task, and this task must be completed only if these n steps are completed consecutively; Each step counts independently of each other; As long as there is a different method taken in one step, the corresponding method of accomplishing the matter is also different.

There are a total of 6 rows. Method 1: The remaining 3 students are assumed to be A, B, and C respectively.

Xiaoli, a, b, c;

Xiaoli, a, c, b;

Xiaoli, C, A, B;

Xiaoli, C, B, A;

Xiaoli, B, A, C;

Xiaoli, B, C, A, there are a total of 6 ways to arrange it.

Method 2: According to the analysis, 1x3x2x1=6 (species) Answer: There are 6 ways to arrange regrets.

Analysis] Xiaoli is fixed in the first place, so there are 3 options in the second place; There are 2 options for the 3rd cavity bucket; There is 1 choice in the 4th place; According to the principle of multiplication, a total of 1x3x2x1=6 (species) can be obtained, and the solution is accordingly.

Four students lined up in a row, with a total of 24 arrangements. 4x3x2x1

12x2x1

24x124 (species).

A: There are 24 ways to arrange it.

Analysis] This question mainly examines the problems of permutations and combinations.

Let the four positions of 4 people be A, B, C, and D. The first position of 4 people can be ranked in the second position, remove the front row of 1 and only 3 people can be left in the third position, remove the front row of 2 people and only 2 people can be ranked in the fourth position, remove the front row of 3 people and only 1 person can be rowed. The equation is:

4x3x2x1 = 24 types of arrangement.

Suppose the first place is called 1, the second place is called 2, and they are arranged in turn, 12341243

1423 is a beginning with 6 kinds, and there are 4 beginnings with 4 times 6

In this question, we need to calculate how many ways to line up with four students, and we need to rule out one case where one of them cannot line up on the left.

First of all, we can consider the number of ways to queue up with four students without any restrictions. For the first position, we can choose any of the four people; For the second position, since there is already one person occupying the first position, there are only three people to choose from; Similarly, for the third position, only two people can choose; And the last spot can only be chosen by one person. To sum up, the number of four students queuing is 4 3 2 1 = 24.

Next, we need to rule out a situation where one of them can't line up on the left. Assuming that the student is A, we can set up A to stand at the top of the line. At this time, the remaining three students in the team can be arranged according to the above method, and there are a total of 3 2 1 = 6 arrangements.

However, we can also put A in any position in the team, so we need to multiply 6 by 4 (A can be placed in any of the four positions) to get a total of 24 ways to arrange that A cannot stand on the left in all cases.

To sum up, when one of the students cannot line up on the left, the number of ways for four students to line up is 24-24=0 rows.

In total, there are 24 ways for the four students to line up. However, if one of them cannot be on the left, then the arrangement is 0. <>

There is the following arrangement:

Xiaojun, Xiaohong, Xiaoling; Brother Songsheng.

Xiaojun, Xiaoling, Xiaohong;

Xiaohong, Xiaojun, Xiaoling;

Xiaohong, Xiaoling, Xiaojun;

Xiaoling, Xiaojun, Xiaoxian Laohong;

Xiaoling, Xiaohong, Xiaojun;

Total: Sakura do 2 3 = 6 (species).

A: There are 6 different types of arrangement

Four students lined up in a row, with a total of 24 arrangements. 4x3x2x1

12x2x1

24x124 (species).

A: There are 24 ways to arrange it.

Analysis] This question mainly examines the problems of permutations and combinations.

Let the four positions of 4 people be A, B, C, and D. The first position of 4 people can be ranked in the second position, remove the front row of 1 and only 3 people can be left in the third position, remove the front row of 2 people and only 2 people can be ranked in the fourth position, remove the front of the Jialing Heng noodles row 3 and only 1 person can be rowed. The equation is:

4x3x2x1 = 24 types of arrangement.

4*3*2*124 types: There are 4 selections for the first position, 3 selections for the second position, and so on

a44, which is the factorial of 4, is equal to 4x3x2x1=24