4 different balls into 4 boxes with different numbers...

1. The first ball can be placed in four boxes.

The second ball can be placed in four boxes.

The third ball can be placed in four boxes.

The 4th ball can be placed in four boxes.

Then there are 4*4*4*4=256 species.

2. Ibid. 3. Empty box No. 1.

The first ball can be placed in 3 boxes.

The second ball can be placed in 3 boxes.

The 3rd ball can be placed in 3 boxes.

The 4th ball can be placed in 3 boxes.

There are 81 types of empty box No. 1.

And there are 4 boxes that can be empty.

Then there are 4*81=324 species.

4. There is exactly one empty box (4 types).

And there is only one box with 2 (3 kinds).

Then there are always 3*4=12 kinds.

5. There are 10 balls that do not need to be changed, and they are placed in the number box.

10 remaining.

1: 4 4 Each ball can be placed in 4 boxes.

3：4*3！4 types of empty box possibilities.

5:4 10 Put enough of the base number of the box first, and then put it as you like.

4 to the 4th power Each ball is freely placed.

Let's look at the box first. Pick two out of 4 boxes, and there are 4!/2!/(4-2)!There are 6 ways to choose.

Let's look at the situation of the ball. The 4 balls are divided into two boxes, 1+3, 2+2, and two divisions:

4 choose 1, 4 kinds, and the remaining 3 choose 3 kinds, 1 kind, combined into 4*1=4 situations;

Choose 2 out of 4 and choose 6 kinds, and choose 2 out of 2 out of 2 and choose 1 kind, which will be combined into 6*1=6 situations.

Combined into 4+6=10 cases.

Overlay calculations. Consider that the order of the group is different, and the scheme is different, so 2*6*10 = 120There are a total of 120 ways to put it.

If there is no special explanation, the default box is generally different, sequential or numbered, and the balls are also different.

There are three major situations to discuss here.

Case 1: An empty box is similar to 0, 1, 2, 3 composed of different four-digit numbers, the difference is that this four-digit number can be 0 in the thousands

There are a total of 4x3x2 = 24 different situations.

Situation 2: If the two boxes are empty, it will be carried out in two steps.

Step 1: The three balls have 3 combinations, namely 12, 13, 23 The second step is similar to 0, 0, 1, 2 to form different four digits, the difference is that this four digits allow the thousand digits to be 0, or the thousands and hundreds to be 0 at the same time

When the thousand digit is 0, there are 3x2x1=6 cases.

When the thousand digit is 1, there are three cases: 002, 020, and 200.

There are also three cases when the thousand digits are 2, so there are 12 different combinations in total.

3x12=36, so case two has 36 combinations.

Situation 3: If the three boxes are empty, there are 4 situations.

Therefore, there are 24 + 36 + 4 = 64 combinations of all the ways to put it.

Is this asking about the result of the box or the order in which it was put in? If it just asks if the box has a ball, it's much simpler:

There is only one case with a ball in one box (4,0,0,0), two cases with a ball in two cases, 2,0,0,,), 3 boxes with a ball in one case (1,1,1,0), a total of 4 cases;

However, if the 4 boxes are numbered, it is complicated: just go in 3 steps:

The first step is to choose a ball among the 3 balls, there are 3 options, and then put the ball into any of the four boxes (such as a b c d), there are 4 ways to put it, a total of 3 4 = 12 results;

The second step is to choose a ball among the remaining 2 balls, there are 2 choices, and then put it into any of the 4 boxes (the title does not say that it should be placed in different boxes), there are also 4 ways to put it, a total of 2 4 = 8 results;

The third step is to put the only remaining ball into one of the 4 boxes, and there are 1 4 results;

Since these three steps cannot be completed independently, it is necessary to multiply them all by all the ways to put the ball, that is: 12 8 4 = 384 kinds of order of putting the ball (it is estimated that it is not what the question wants).

1) Put the first ball first, you can put 4 boxes, there are 4 options.

Put the second one, there are also 4 options.

By analogy, through the principle of multiplication, there are a total of 4*4*4*4=256 methods.

2)a.First, 2 empty boxes were selected, a total of 4c2=6 methods.

b.Place the balls in the remaining 2 boxes:

1) 1 piece in 1 box, 3 pieces in 1 box: first grouped, then arranged. (4c1*3c3)2p2=8 species.

2) 2 pcs per box: 4c2*2c2=6 kinds.

1) 2) is the principle of addition, AB is the principle of multiplication.

Therefore, the answer to this question is 6*(8+6)=84 kinds.

3)a.Put C and D first, the same as (1), a total of 4 2 = 16 methods.

b.Then put A and B.

1) If A puts 1 box, then B can put the box. 3 types in total;

2) If A puts 2 boxes, then B can put boxes, a total of 2 kinds of inspection;

3) If A puts 3 boxes, then B can put 4 boxes, a total of 1 kind.

1) 2) 3) is the principle of addition, ab is the principle of multiplication.

So the answer to this question is 16*6=96 kinds, 3,1Each ball has four boxes to choose from, so it's 4*4=16

2.That is, put four balls into two boxes, which is c43 + c42 + c41 = 83That is, first consider the placement of the two balls of A and B, and then consider the placement of the remaining balls, so as not to give the answer, 0, four different balls, all put into the four boxes numbered 1, 2, 3, 4.

1) How many ways to put it casually (there can be empty boxes, but the balls must all be put in the box?).

2) How many ways are there to put exactly two empty boxes?

3) How many kinds of coarse and quiet ways for the number of the box placed by ball A is always smaller than the number of the box placed by ball B?

Each of the 16 balls can be placed in 4 boxes numbered 1234. There are 4 cases. That is, each ball can be placed in 4 different boxes.

And because there are 4 identical balls. So there are 4*4=16 ways to put it in total.

There are four ways to place each ball: put in the first, put in the second, put in the third, and put in the fourth box. There are four balls in total, so 4*4*4*4=256 ways to put it.

4*4*4*4*4+c51*3*3*3*3+c52*2*2*2+c53

At the same time, each box can hold 4 different numbers One is the same has five kinds, the last four, each has 3 ways to mount, 2 have C52 when they are the same, and the last three have 2 kinds of each of you When the three are the same, there is C53

Four identicals are five identical.

Put one in each box, there are 4!Planting and release method.

There are 4 ways to put a box empty: c(4,1)*3.

The answer to the first question is: 4*3*2*1=24

The answer to the second question is: 3*3*2*1=18

First: Since the balls are the same, and the number of boxes is the same as 4, therefore, there is only one possibility to put one in each box, and four balls are placed casually, as long as there is one in each box, the effect is the same (but if the balls are different, the result is not 1), so there is only one way to put it.

Second: First of all, choose this empty box, there are four possibilities. Then put four balls into three boxes (at least one ball in each, so put three balls into three boxes first, one for each of them), and then choose one ball from the three boxes to put in, a total of three possibilities, so there are a total of 4*3=12 ways to put it.