The principle of high school counting is evenly grouped

Divide into n groups, then divide by n!

For example, if you divide 3 books into 3 groups, it is clear that there is only one way to divide them.

And we get a result of c(3,1)c(2,1)c(1,1).

Consider numbering the three books as A, B, and C

c(3,1)c(2,1)c(1,1) can be a,b,c,a,c,c,b,b,a,c,c,c,a,a,c,a,a,b,c,a,b,c,a,a,b,c,aThis is equivalent to sorting a, b, and c, so divide by a(3,3)=3!

Therefore, there is the above conclusion: divide by n!

Suppose the six books are A, B, C, D, E, and F

c 62 means to take any two of the above six numbers;

c Forty-two means to take two of the remaining four numbers;

c 22 means take any two of the remaining two numbers;

The result is that the following six groups are actually exactly the same, but you see them as different groups in the above process, and the factorial divided by 3 is actually divided by the number of repetitions.

c six two c four two c two two.

a,b c,d e,f

a,b e,f c,d

c,d a,b e,f

c,d e,f a,b

e,f a,b c,d

e,f c,d a,b

The combined sequence is out of order, but there is a process of arranging after you divide it, and the arrangement is orderly.

As a simple example, there are two numbers, and the average is divided into two groups, according to your understanding, C21C11 = 2 kinds, but no matter how you divide them, they can only be a group, B group, there is only one situation, why?

Because when you are in C21, it may be A, and then C11 is B, or C21 is B, and C11 is A, so there is A, B, B, A, and then it is actually a permutation number, and the problem only requires grouping and does not require permutation, so it is divided by the factorial of the number of groups.

Questions 1 and 2 are not the same problem.

Question 1: c6(2)*c4(2)=90

Question 2: c6(2)*c4(2) (3*2*1)=15Question 3: Equivalent question 1: c6(2)*c4(2)=90

Solution:1First, take 2 copies from 6 books to A, then take 2 copies from 4 books to B, and the remaining 2 books to C, a total of C2,6*C2,4*C2,2=90 kinds.

2.For the problem of equal stacking, first take 2 books from 6 books to make a pile, and then take 2 books from 4 books to make a pile, and the remaining 2 books are a pile, but it should be noted that when the average pile is divided, it should be divided by the number of full arrays of the number of piles, and the uneven stacks do not need to be divided, total.

c2,6*c2,4*c2,2 a3,3=15 kinds of 3Without distinguishing between people, it can be seen as an average score of 3 parts, which is equivalent to question 2, which is also 15 types.

Hope to understand.

Question 1: 6 different books are divided into 2 copies for each of the 3 people A, B and C, how many ways are there to divide them?

c(6,2)*c(4,2)*c(2,2)=90Question 2: 6 different books are divided into 3 parts, 2 copies each, how many divisions are there?

c(6,2)*c(4,2)*c(2,2)/a(3,3)=90/6=15

Question 3: 6 different books are divided into 3 people, 2 books per person, how many divisions are there?

Questions 1 and 2 of the same question are obviously not the same question, because the answers are different

1. C2, 6 * A1, 3 * C2, 4 * A1, 2 = 540 2, C2, 6 * C2, 4 = 90

3. C2,6*A1,3*C2,4*A1,2=540 Questions 1 and 2 do not belong to the same problem.

One: First divide the six books into three equal parts, and give them to the three people: 15x6x1x6 = 540 kinds of two: six books are divided into three parts: 15x6x1 = 90 kinds of three: 540, which is about the same as one.

One is different from two, because one has to take into account that different people take different books, while two is just divided into three parts, and there is no need to think about anything else.

Question 1: C62C42C22 A33 multiplied by A33 (C62 is to choose 2 out of 6, I think you understand) It belongs to the problem of grouping first and then assigning The book needs to be divided by A33 after dividing it into three groups, because it is evenly grouped, remember to divide it by a few groups equally, and then multiply A33 by A33 to divide the three groups of books into three people.

Question 2: It is a grouping problem, just c62c42c22 a33.

In addition, questions 1 and 2 are not the same problem, while questions 1 and 3 are clearly the same problem.

,2*c4,2*=90

3.Isn't it the same as the first?

1. There are 15 combinations of 2 out of 6 groups, 6 combinations for 3 people, and 90 kinds of multiplication.

2. 15 species.

3. If you don't distinguish between people, there are 15 types.

12 books divided into 2:2:2:6 probability: c12,2 * c10,2 * c8,2 * c6,6 = 83160

6 Possibilities of Books in Whose Hands: 4

(c126*c21*c62*c42)*a44/(a22*a33)=332640

Among them, C126 represents 6 out of 12.