There are 3 different books divided among 3 children, how many divisions are there?

Divide 3 books into 3 students, one for each person, and there are 6 ways to divide them. Solution: According to the meaning of the title, it can be seen that this title is the full arrangement of the three books.

then p3 = 3 * 2 * 1 = 6 (species). If the three books are A, B, and C, and the three students are A, B, and C, then the specific 6 divisions are as follows. 1. A is divided into A books, B is B books, and C is C books.

2. A is divided into A books, B is C books, and C is B books. 3. A is divided into B books, B is a book, and C is C books. 4. A is divided into B books, B is C books, and C is C books.

5. A is divided into C books, B is B books, and C is C A books. 6. A is divided into C books, B is divided into A books, and C is divided into B books. 1.

Addition principle: to do one thing, to complete it can have n types of methods, in the first type of methods there are m1 different methods, in the second type of methods there are m2 different methods ,......There are mn different methods in the nth type of approach, so there are n=m1+m2+m3+....+mn different methods. 2.

The methods of the first type of approach belong to set A1, and the methods of the second type of approach belong to set A2,......The method of the nth type of method belongs to the set an, then the method of accomplishing this belongs to the set a1ua2u....Requirements for classification: each method in each category 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). Permutation a(n,m) = n (n-1).

n-m+1）=n!/（n-m）!(n is the subscript, m is the superscript, the same below) combination c(n,m) = p(n,m) p(m,m) =n!

m!（n-m）!；For example, a(4,2)=4!

2!=4*3=12c(4,2)=4!/(2!

There are 6 ways to do this. What grade are you asking this question in, if it is a high school direct permutation and combination, a 3, that is, 3x2x1 6

If it's a lower area, it's like sorting it, 123,321,132,312,213,231. That's it, of course, only for small quantities.

There are 3 different books, divided among 3 children, and there are 6 divisions.

Permutations and combinations, if each child is asked to have a copy, there are: 3! = 6 divisions.

In this case, there are 6 divisions.

Teacher, there are four identical books, and they have to be divided among the three children, that is, one for each of Qingyuan, so do you want one more for one child? So his division depends on whether you ask everyone to get how much, or to divide it all, that is, everyone can get a copy, that's the kind of thing, if you think you have to give the rest of the book to the children. Then there is more envy of this division.

The four books are numbered as ABCD, the three children are 123, and there may be four kinds of books for the students of the No. 1 Banquet, if so, the good students may get three kinds of books, and the third student has two, then all possible situations are 4 3 2 = 24 kinds.

Teacher, having four identical books should be distributed to three children, that is, one book for each child. Do you want to buy an extra book for each child? Therefore, the division of its premature sensitivities depends on how much you want everyone to get, and Lu Wuzhi still wants everyone.

Inside the nuclear stupidity.

1.The first way to divide is to give the first child a story book, the second child a popular science book, and the third child to accompany a picture book.

2.The second division is to give the first child a popular science book, the second child a picture book, and the third child a storybook.

3.The third division is to give the first child a picture book, the second child a story book, and the third child a popular science book.

4.The fourth division is to give the first child a storybook and a popular science book, the second child a storybook and a picture book, and the third child a popular science book and a picture book.

Summary. There are three different books, one for each of the three children, and there are many ways to give them.

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6 kinds. Assuming that the numbers of the three books are E, F, and G, there are 6 ways to send them: EFG, EGF, GEF, GFE, FGE, and FEG. According to the knowledge of permutations and combinations, first send a book, there are 3 ways to send, and then send a second book, at this time only two children have no book, then there are 2 ways to send, and finally there is only 1 book and 1 child left, then 1 way to send.

So, a total of 3x2x1 = 6 (kinds) of delivery.

13 3 = 4 (Ben)....1 (Ben), 4+1=5 (Ben);

A: At least one child at the age of closure can get 5 books

So the answer is: state acacia

There are 6 different books, divided among 3 students, two each, and there are several divisions.

Hello, according to the problem you described, there are 6 different books, divided into 3 students, 2 books each, there are 90 divisions. The first person chooses two books, there are 6*5 2 kinds of choices, the second person chooses two 4*3 2 books in the remaining 4 books, and the last one has a total of 6*5*4*3 4 kinds. The above is for your reference.

There are a total of 6 ways to deliver it.

Assuming that the numbers of the three books are A, B, C, and hail dust, there are 6 ways to send them: ABC, ACB, BAC, BCA, CAB, and CBA.

According to the knowledge of permutation and combination, first send a book, there are 3 ways to send, and then send the second book, at this time there are only two children without a book, then there are 2 ways to send, and finally there is only 1 book and 1 small friend missing to search for friends, then 1 way to send. So, a total of 3x2x1 = 6 (kinds) of delivery.

Definition of permutation and its calculation formula:

Taking m elements from n different elements in a certain order is called taking out an arrangement of m elements from n different elements. The number of all permutations of m (m n) elements from n different elements is called the number of permutations of m elements from n different elements, a(n,m) = n(n-1)(n-2) ......n-m+1)= n/(n-m) 。

The definition of the combination and its calculation formula: from n different elements, take any m(m n) and return the elements to form a group, which is called a combination of m elements from n different elements; The number of all combinations of m (m n) elements taken out of n different elements is called the number of combinations of m elements taken out of n different elements. It is represented by the symbol c(n,m).

c(n,m)=a(n,m)/m。