A question about the permutations and combinations of the book

9 books, divided among three people, three books for each person, ask how many divisions there are.

The correct answer is c(3,9)*c(3,6).

Why can't it be c(3,9)*c(3,6)*a(3,3)?

Why can't you just split it up and then arrange it and send it down?

Because for you: the first, the second, the third and to you: the second, the first, the third.

These two cases are the same, that is, the order does not need to be considered, so instead of *a(3,3)9 books, it is divided among three people, one person has four books, one person has three books, and one person has two books.

The correct answer is c(4,9)*c(3,5)*a(3,3), why can't it be c(4,9)*c(3,5)?

Why do you have to divide the piles first, then arrange them, and send them down?

Because the first person takes four copies and the second person takes four copies is not the same, that is, the order must be considered. Hence the need for *a(3,3).

If you don't think it's clear enough, just ask.

Wish: Learning progress!

1: There is no order relationship between the three people, the first one takes 3 copies, the second one takes three copies, and the rest is finished.

2: Because the number of people to take is different, there are 6 kinds of sequential relations, so it is c(4,9)*c(3,5)*a(3,3).

1) There are three places to go for each book, so it should be 3 to the 5th power. i.e. 243 species.

2) This is a permutations and combinations, and there are 150 answers. If you want a detailed answer, you can ask me for help.

1) 5 different books each have three places to go, so it is 3*3*3*3*3=3 5=243

2) Everyone has at least one copy, there are only two ways to divide the pile, 113 or 122:, and the two methods of dividing the pile are calculated separately below.

Note: A32 refers to 3 in the lower 2 in the top).

113: The first step: pick one first, and then pick one of the rest to complete the pile, that is, 1 book, 1 book, 3 books, c51*c41*c33

Step 2: Assign these three piles to 3 people, a33, because two of them have the same number to produce duplicates, so divide by 2

The final formula is: c51*c41*c33*a33 2

122: The method is the same as above, and the final formula is: c51*c42*c22*a33 2

Finally, add the two up and it comes out to be 150

Because there are duplicates in the uniform grouping, the possibility of duplicates should be ruled out.

There are no duplicates in the uneven grouping, so there is no need to rule them out.

In the example given, a,bc,def and def,bc,a are grouped together, and in the calculation of c(6,1)*c(5,2)*c(3,3), there is only one component at all.

Ab, Cd, EF and Cd, Ab, EF (and the other 4 kinds) are counted as different groups in the calculation of C(6,2)*C(4,2)*C(2,2), when in fact the 6 of them are only used as a group.

That is, the calculation of c(6,2)*c(4,2)*c(2,2) actually contains the components of the permutation (the permutation is added invisibly in the step).

c2 6c2 4c2 2 is a formula that divides 6 different books into 3 parts, and arranges and combines these 3 parts. However, the question does not require dao to permutate and combine the 3 parts, so divide by the number of permutations and combinations: a3 3.

If the title is changed to "6 different books, evenly divided among 3 students", there is no need to divide A3 3.

It's easier to fix three people and think about them with five different books one after another.

In the first book, there are 3 divisions;

In the second book, there are also 3 divisions;

Up to the fifth book, there are three options, and when the fifth book is allocated, all the allocations are completed.

The answer should be: 3*3*3*3*3=243

There are three people in A, B and C, and five books in ABCDE.

Give A first, there are five books to choose from.

In giving to B, there are four books to choose from (A gets one).

Finally, to C, there are three books to choose from (for the same reason).

Therefore, it is 5*4*3=60

Upstairs is also available, and the people are arranged as A33 first

Divide the book again, for C53, but I.

The method of speaking should also be understood.

1) There are three places to go for each book, so it should be 3 to the 5th power. i.e. 243 species.

2) This is a permutation and combination question, bluffing and slackening, and the answer is 150 kinds. If you want a detailed answer, you can ask me for help.

The first person has five choices, the second person has four choices, and the last person has three choices, all answers are: 5*4*3=60, i.e. a53 or c53a33can be used as an answer!

There are three people in A, B and C, and five books in ABCDE.

Give A first, and there are five books to choose from.

In giving to B, there are four books to choose from (A gets one).

Finally, for C, there are three books to choose from (the reason is the same as above), so it is 5*4*3=60

Upstairs is also available, and the people are arranged as A33 first

Again, it's C53, but the method I'm talking about should still be understandable.

Because there is no sequential issue involved, just simple grouping, combinations are used.

Suppose these 6 are 1, 2, 3, 4, 5, 6

First choose two of these 6 books to form a group, a total of 62 types, assuming that 2,3 is selected, then choose two of the remaining 4 books (1456) to form another group, a total of 42 kinds of C, if 1,5 is selected

In the end, there are 2 books left (46) that are not chosen, so you can only choose these two books to form the last group, a total of 22 types.

So there are a total of c62*c42*c22=90 kinds.

If the group is numbered, then do it with permutation.

Of course not!

Divide 5 different books into 3 people" emphasizes different books, e.g. I give book 1 to A and book 2 to B - compared to I give book 1 to B and book 2 to A, these are two different situations. This emphasizes both the number of books and the different categories of books. So it's a matter of arrangement.

Divide 5 books among 3 people"If there is no manual, the books may or may not be the same, but this is not the focus of his emphasis, what he emphasizes is only the number of books that each person gets, and it can be divided. So it's a combination problem.

If you read these two sentences carefully and see the point of what he said, it should not be difficult to understand.

It's all a matter of combination, because three people are different, and there is an order, but each person gets a book that is not in order, and it is a combination, but when it is divided, it has been divided into three people in order, so the rest is a combination of books.

Because,,The amount of books that 3 people get is different,The first question is the average score,Each person is two,There is no consideration of the arrangement on 2,2,2,2,In one case, note that only the number of books is considered here, and nothing else is considered.

The second question is not an average score, there are 1,2,3,,, 1,3,2,,,2,1,3,,2,3,1,,,3,1,2,,,3,2,1 The number distribution of these six different books must be multiplied by a33,