20 identical balls in 3 boxes numbered 1 2 3 Question 10

It is done with the partition method, which is the same as the insert method on the third floor.

The first question is to divide the 20 balls into 3 parts, as long as there are 2 boards, so it is c(19,2), not c(19,3) as said on the fourth floor, and it is divided into 4 parts according to the fourth floor.

In the second question, you first put a ball in the box numbered 2 and 2 balls in the box numbered 3. In this way, there are only 17 balls left, and then the 17 balls are divided into 3 parts by the partition method, which is c(16,2).

A little more troublesome can be used with the plug-in method.

2)+c(18,1)=171

2)+c(15,1)=120

There's also a little stronger column:

The idea of imaginary addition is mainly used.

Fill in one ball first, then insert two versions.

Remove one more ball from the two boards.

and the above is equivalent.

The answer to 1 is to divide 20 balls into 3 groups with a wooden plank, that is, c(19,2)=19*18 2

The answer to 1 is to divide 20 balls into 3 groups with a wooden board, that is, c(16,2)=16*15 2

19*18*17= 215*14*13= The specific data is calculated by yourself....

Because the number is greater than the number, you must put 2 in the first box, 3 in the second box, 4 in the third box, and the remaining 6 balls at will. Divided into 6 and found for 3 groups, respectively, 006,015,024,033,114,123,222, a total of 7 kinds and then arranged in one, two, three boxes, so there are 3 + 6 + 6 + 3 + 3 + 6 + 1 = 28

55 kinds, first put one ball in the second box, put two balls in the third box, and then partition the plate, c2 11 = 55

The original question is equivalent to putting 7 balls into 3 boxes, with at least one ball in each box, then adding 1 ball in the second box and 2 balls in the third box.

In this way, you can use the "insertion method": arrange 7 balls in a row, insert 2 or 2 "boards" in the middle of 6 gaps, and divide the balls into 3 piles, so as to obtain a division. So there are a total of c(2,6)=15 methods. #

15 kinds. Put a small ball in each box first, so as to ensure that the number of small balls put into it is not less than the number of the box;

Put the remaining four balls into three boxes and you're done.

The number of balls is not less than the box number, that is, the No. 1 bottle is filled with at least one ball, the No. 1 bottle is filled with 1 ball, the No. 2 bottle is filled with two balls, and the No. 3 bottle is filled with 3 balls.

There are 4 balls left, each with three slots to play.

Three to the fourth power.

81 species.

First, put numbered balls in each box, i.e. box 1 for 1 ball, box 2 for 2 balls, and box 3 for 3 balls. There is 1 type in total.

The remaining 14 balls can be placed casually, and they can be divided into 3 groups by interpolation, with a total of 16 empty positions, so there are c(16,2) kinds.

So there are a total of: c(16,2) = 120 species.

The three boxes are placed 1, 2, and 3 respectively, and there are 14 left to be placed casually, which is equivalent to inserting 2 partitions in 17 positions, a total of 16c2=120 kinds.

Method 1: Put 1 and 2 balls in ball 2 and 3 respectively, then there are 17 balls left, and the problem is transformed into:

Put 17 balls in three boxes, each box has at least 1 ball, how many kinds are there?

Typical "baffle method" problem!

17 balls are lined up in a row with 16 gaps, and 2 baffles are inserted.

c(16,2)=120

Method 2: According to the title, first put 1 ball in the box numbered 2, put 2 balls in the box numbered 3 in turn, and there are 17 balls left, just put the 17 balls into 3 small boxes, at least one for each small box, a total of 16 empty spaces between the 17 small balls, choose 2 of them, insert the baffle, then there are C162 = 120 different ways to put it, so the answer is: 120

According to the title, a small ball is placed in each of the three boxes numbered at first, and the box numbered 1 is not placed; Put the remaining 7 balls into 3 boxes, at least one in each box; Total: c2

2 1=15 (species);

There are a total of 15 ways to put the questions that meet the requirements of the question

So the answer is: 15

Or another solution: There are five ways to put box No. 1: 1, 2, 3, 4, 5 Talk about it separately, when box No. 1 is a ball, first row box No. 2, a total of 2, 3, 4, 5, 6 five ways to put When there are two boxes No. 1, there are four ways to put it, and so on, a total of 5 + 4 + 3 + 2 + 1 = 15 ways to put it

You're right to think so, but 14 3 you're not on the same page as the right one. 3 to the 14th power I can still understand, 14 3 is what's going on?

Even if it is 3 to the 14th power, there is still a problem, because there are duplicates, and there are a lot of repetitions, very many ......

The correct solution is the interpolation method.

Suppose 14 balls are lined up in a row, and the 14 balls have a total of 13 gaps, plus the two ends, there are 15 spaces, which now translates into the number of permutations with three lockets inserted into 15 gaps. The correspondence is: to insert.

The number of balls between the two empty boxes indicates the number of balls in the lockets on the right space, and the leftmost space can be inserted into two lockets at the same time. The rest of the gaps can only be inserted into a small box, the rightmost space must be inserted into the box, if there are two lockets inserted into the leftmost space, yes.

c(2,3) species; If there is exactly a cascadet inserted into the leftmost space, there are c(1,3)c(1,3) kinds;

If there is no caslet inserted into the leftmost space, there are c(2,13) kinds, by the principle of addition, there is.

n=c(2,3)+c(1,3)c(1,3)+c(2,13)=120 permutations, i.e., there are 120 ways to put them.

It's not right to think that way.

What if the whole part goes to the first box, or the whole part goes to the second or third box?

Step 1: Each middle is the same number as the number: 1 + 2 + 3 + 4 = 10 Step 2: The remaining two, two groups, there are four ways to put them, and the two are separated: c(4,2)=6 (species).

So there are a total of: 4 + 6 = 10 (species).

10 kinds First of all, the number of the question requirements can not be less than the number of balls, so 1 2 3 4 The four boxes add up to at least 10 balls, and there are two balls to put in the four boxes There are two ways One is C42 and one is C41 A total of 10 kinds.

This problem uses the partition method, but it needs to be transformed in one step.

If three boxes are respectively filled with A, B, and C in the lifting shed, then A+B+C=20, and A is greater than or equal to 1, B is greater than or equal to 2, and C is greater than or equal to 3.

Let x=a, y=b-1, z=c-2, then x, y, and z are all greater than or equal to 1 (this is the condition of the partition method).

So x+y+z=17

The problem translates into placing 17 balls in three boxes, at least one for each gently wide box, using the partition method. That is, 17 balls are lined up in a row, and two boards are placed in the middle, and the total number of boards is c16,2.

Answer: c16,2 = 120