Six excellent league members will serve as counselors in Class 7 1, Class 2, Class 3 and Class 4, and each class will have at least 1 students. What are the different distribution methods? What is the

Six excellent league members will serve as counselors in Class 7 1, Class 2, Class 3 and Class 4, and each class will have at least 1 students. What are the different distribution methods? What is the correct way and answer to this question? Good, safe.

The best and clearest solution is

1. First, choose a class from four classes, so that three people can have C (6 6,3 3) A (4 4,4) = 480 kinds.

The idea is that three of the six people are "tied" together and regarded as an element. It is completely consistent with the other three elements. Consistent with your and his answers.

2. Choose two classes from four classes, with two people in each class, and one person in each of the other two classes C (6,2) C (4,2) A (4,4)/2! = 1080 species.

The idea is that two of the six people are "tied" together and regarded as an element. Two of the four people are "bound" together and regarded as one element. Exactly the same as the other two elements.

Consistent with your answer. The reference book is wrong.