How many ways to put 10 identical children in 4 different rooms?

The first question starts with the fact that there are no children in the first room.

There are 11 types at 0,0,x,x (from 00,0,10 to 00,10,0).

10 at 0,1,x,x.

There are 9 types of 0,2,xx all the way up to one of 0,10,00.

The first room without kids has a way (11+10+......1) = 66 species.

The first room has a child from time to time (10+9+......1) = 55 species.

Then there are 45, 36, 28, 21, 15, 10, 6, 3, all the way up to a total of 10,000 (66 + 55 + 45 + 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1) = 286.

The second question is equivalent to 6 children in 4 rooms, because there is at least one in each room, and there can be no more than 4 in one room of 6 children.

There are 19 ways for one child in the first room, 18 for two children, 15 for three children, 10 for four children, and 6 for five children, for a total of 68 types.

I always feel that there is a simple algorithm for the second question, so let's talk about it after thinking about it; In addition, thanks to your list of answers, I got it wrong 800 times to get it right, especially the second question, which is easy to miss.

1,2,3,4 and 2,2,3,3, the former has 24 ways:

There are 6 types of the latter:

There are 30 types of one in total.

10 children in 4 rooms: combination coefficient: c up 4 down 10 = 210 types; At least one person:

210-C up, 3 down, 10-c, 2 up, 10-c, 1 up, 1 down, 10 = 35 kinds, more than 5 people: C up, 3 down, 4 = 4 kinds, so there are 35-4 = 31 kinds of children in each room and no more than 5 people.

There are four ways for each child to put it, that is, 4 to the 10th power.

Group discussion, first, 10 people together, there are a41 = 4 kinds of release.

Second, when divided into two groups, (9,1)(8,2)(7,3)(6,4) each group has a42(4*2)=12 releases, for a total of 48 releases. When both groups are 5 people, it is halved and there are 6 types of hair.

Third, when divided into three groups, (7,2,1)(6,3,1)(5,4,1)(5,3,2) each group has a43(4*3*2)=24 methods, a total of 24*4=96, (8,1,1)(6,2,2)(4,4,2)(3,3,4) halved in each group, a total of 48 releases.

Fourth, when divided into four groups, (1,2,3,4) has a44(4*3*2*1)=24 methods, (1,1,2,6)(1,1,3,5)(1,2,2,5) halved a total of 36 methods, (1,1,4,4)(2,2,3,3) has c42(4*3 2)*2=12 methods, (1,1,1,7)(2,2,2,4)(3,3,3,1) has 3*a41=12 methods.

The above total (4 + 48 + 6 + 96 + 48 + 24 + 36 + 12 + 12) = 286

In the second question, subtract (1,1,2,6) from the 12 (1,1,1,7) 4 types in the fourth question above, totaling 84-12-4=68

Put all the children in one room + choose two rooms and then use the partition method, and then use the partition method for two + choose three rooms and then use the partition method + 4 rooms and then use the partition method to ......This is 4+c4 2 * 9+c4 3 *c9 2+c9 4=296......I suspect your answer is wrong......Hehe.

On average, there are two in each box.

Then load the remaining two balls in turn according to the rules.

The result is 10 kinds.

I probably did a year ago.

I would have done it a year ago...

First, each box is filled with their (number of boxes - 1) balls. After this is done, there are 4 small balls left. Then we need to use the partition method.

First, line up the remaining 4 balls in a row (casually) so that there will be 3 spaces in the middle of the balls. Then we can separate from each other by bounding it. In addition, the small ball in the original box just meets the requirements of the question.

In the calculation process, start with 3 out of 3 in the air. The C 3 method should be satisfied.

6 balls are the same, 4 boxes are different, and the difference is in the box.

Since each box is not empty, first take out 4 balls and place them in 4 different boxes.

Solve this problem.

First of all, fill all 4 boxes.

There are 4 ways to do this.

The other two have C4 2

Species = 2x3 = 6 species.

So there are altogether.

6+4=10 species.

Hope it helps.

I don't know how to ask questions.

I'll help you answer, remember to adopt it.

The number of balls in each box is denoted as x1, x2, x3, x4, and it can be seen that a release method corresponds to a non-negative integer solution of the equation x1+x2+x3+x4=7.

Therefore, the number of ways to release depends on how many sets of non-negative integer solutions there are in the equation.

Suppose there are 10 pebbles lined up, choose any 3 of them to mark, and these three marked stones divide the remaining 7 pebbles into four parts, and it can be seen that one selection corresponds to a set of solutions of the equation x1+x2+x3+x4=7, so the answer is c(10,3)=(10*9*8) (3*2*1)=120.

There are two ways to put the same pit, one is 6 1 and two 2 (the size of the pit is just 2), a total of 28 ways; The second is 7 1 and a 3 (the size of the pit is just 3), a total of 8 ways to put it.

Different pits can only be placed in 1, 2, 3, 4. There are a total of 24 ways to put it.

There are 60 ways to put it in total, and I don't know if it's right or wrong.

This kind of combinatorial math problem of solving the number of schemes has a formula, model:

n different balls.

m different boxes.

Number of balls per box" = 1

There is this article in the library.

Various situations have been discussed, if only the number of schemes is required, you can directly set the formula and compile a few functions of the factorial.

There may be an easier and faster way, here is the traversal result: 84 include

using namespace std;

int main()}

cout<

First, put 1 ball in each box, and then the remaining 6 balls can be placed as many as you like.

How many ways to put 10 identical balls in 4 different boxes, and at least 1 ball in each box?

6 identical balls in four different boxes, if there is 1 in 3 boxes and 3 in 1 box, there are 4 ways to do this;

If there are 2 boxes in 2 boxes and 1 in 2 boxes, in this case: let the four boxes be numbered as The case of the box with two balls may be as follows: So there are 6 cases;

Therefore, C