Put two people in the front row. Then choose one of them first, with C 12 possibilities, and insert this person into four people, with A 15 possibilities. (Because there are five vacancies in * * *, the same below)
If you insert another person into five people, there are 16 possibilities
So the answer is: C28 * c12 * a15 * a16 = 28 * 2 * 5 * 6 =1680.
2. If there is at least one student in each class, three students will be selected first. There is a possibility of C34. There are 33 possibilities for inserting into a class.
The fourth student is randomly inserted into three classes, with A 13 possibilities. Therefore, four students can be assigned to three classes, and each class has at least one student: C34*A33*3.
Considering that A can't be classified into Class A, consider the possibility of A being classified into Class A first. .
A In A, the other three races took out two people to join the other two classes. * * * There is the possibility of C23*A22. There are three possibilities for the fourth person to randomly insert into three classes. So * * has C23*A22*3.
So there is a * * * scheme that meets the requirements of the topic: C34*A33*3-C23*A22*3=54.
The answer is 54.