Five balls in seven boxes asked

1.There are 7 5 ways to put (7 5 is 7 to the 5th power) because each ball has 7 ways to put it, 5 balls have 7 * 7 * 7 * 7 * 7 = 7 5 (kind) 2The probability of five balls appearing in a box at the same time is 1 7 4, because there are 7 cases when five balls appear in a box at the same time, divided by the total number of cases, it is 7 7 5 = 7 4

3.The probability of up to four balls in the same box is 1-1 7 4 because the pros of up to four balls in the same box are at least five balls in the same box, and the probability of five balls in the same box is 1 7 4, so the probability of up to four balls in the same box is 1-1 7 4

4.The probability of the same box of up to three balls is 1-16 7 4 First count four balls in the same box, the number of cases is 5 * 42 = 210, and the probability is 210 7 5 = 30 7 4

The probability of five balls in the same box is 1 7 4

Therefore, the probability of at least 4 of the same box is 30 7 4 + 1 7 4 = 31 7 4, so the probability of his opposite, that is, the maximum of three balls, the probability of the same box is 1-31 7 4 behind. It's too hard to express.

It will be easier to write with a pen.

There are some symbols on the computer that can't be typed. I'm sorry.

You can make the five balls a1a2a3a4a5 and the seven boxes b1b2b3b4b5b6b7, 1) five balls in the same box: 7 (7 5) = 1 (7 4);

2) Up to one ball in the same box: (7*6*5*4*3) (7 5)=360 (7 5);

3) Up to four balls in the same box: probability of 1-5 balls in the same box = 1-1 (7 4);

4) Up to three balls in the same box: the probability of up to four balls in the same box - the probability of four balls in the same box = 1-1 (7 4)-5*(7*6) (7 5) = 1-1 (7 4)-30 (7 4) = 1-31 (7 4);

5) Up to two balls in the same box: the probability of one ball in the same box at most + the probability of two balls in the same box = (7 * 6 * 5 * 4 * 3) (7 5) + (5 * 3) * (7 * 6 * 5) (7 5) = 360 (7 4) + 450 (7 4) = 810 (7 4).

6) The probability of the box appearing without putting the ball: 1, that is, the probability of appearing with a box without putting the ball is 1

No, the probability of the box where the ball is placed is 0, that is, the probability that there is no ball in a box is 0

Note: The content in {} is an explanation of the previous data.

What is the probability that five balls will appear in a box at the same time.

There are no probabilities.

Give me some points and I'll do it.

After reading the theory of probability, this one is simple.

None of the above is right.

Summary. Dear, hello, according to the meaning of the question can be solved: assuming that the three boxes are different boxes, the solution idea:

Pick the ball + put the box. After classification and discussion, there are the following two situations: 1) 5 = 3 + 1 + 1, that is, there is a box with 3 balls, and the other two boxes each put 1 ball, so there are a total of c 3 1 * (c 5 3 * c 2 1 ) = 3 * 10 * 2 = 60 kinds; 2) 5 = 1 + 2 + 2, that is, there is one box for 1 ball, and the other two boxes each put 2 balls, so there is a total of c 3 1 * (c 5 1 * c 4 2 ) = 3 * 5 * 6 = 90 kinds; To sum up, 5 different balls into 3 identical boxes, each box with at least one ball, a total of 60 + 90 = 150 different ways to put it.

5 different balls into 3 of the same box, each box with at least one ball, how many ways to put it.

Dear, hello, according to the meaning of the question can be solved: assuming that the three boxes are different boxes, the solution idea: ball selection + box.

After classification and discussion, there are the following two situations: 1) 5 = 3 + 1 + 1, that is, there is a box with 3 balls, and the other two boxes each put 1 ball, so there are a total of c 3 1 * (c 5 3 * c 2 1 ) = 3 * 10 * 2 = 60 kinds; 2) 5 = 1 + 2 + 2, that is, there is one box for 1 ball, and the other two boxes each put 2 balls, so there is a total of c 3 1 * (c 5 1 * c 4 2 ) = 3 * 5 * 6 = 90 kinds; To sum up, 5 different balls into 3 identical boxes, each box with at least one ball, a total of 60 + 90 = 150 different ways to put it.

The same box.

Dear, the answer is the same.

The answer is as follows: Since there are five balls in total, one per box on average, one ball is placed in each box.

There are a total of 5 balls, how many balls are placed in each box?

c（5，1）

1 of you is not aggressive.

c（5，2）

Hopefully my key to you contains the help of the manuscript Zen laugh!

243 species. Because each ball has 3 options, it is 3 to the fifth power, 343 kinds.

Permutation: From n different elements, any element of m (m n, m and n are natural numbers, the same below) elements are arranged in a certain order, which is called an arrangement of m elements from n different elements; The number of all permutations of m (m n) elements from n different elements is called the number of permutations of m elements from n different elements, which is represented by the symbol a(n, m).

Calculation formula: combination: from n different elements, take any m (m n) elements and form a group, which is called a combination of m elements from n different elements; The number of all combinations of m (m n) elements taken out of n different elements is called the number of combinations of m elements taken out of n different elements. It is represented by the symbol c(n,m).

Calculation formula:

1.Different balls have different boxes, and for balls, each ball has 3 choices, so it's 3 to the fifth power.

2.For the same ball and the same box, there are 500, 410, 320, 311, 221 by enumeration.

3.Different balls in the same box, according to the above 5 cases to find the combination and then sum, there are 1 + 5 + 10 + 25 + 30 = 71 kinds.

4.Different boxes for the same ball, there are 3+6+6+3+3=21 types.

4+4+12=20 (species) Answer: There are 20 different ways to put it. Method:

1,1,1,4 1,1,4,1 1,4,1,1 4,1,1,1 1,1,2,3 1,1,3,2 1,2,3,1 1,3,2,1 1,2,1,3 1,3,1,2 2,1,1,3 2,1,3,1,1,3 2,1,3,1,1,1,2 3,1,2,1,1 3,2,1,1 1,2,2,2,2,2,2,2,2,2,2,2,2,2 It is easy to get the following three cases: 7=1+1+1+4 7=1+2+2+2 7=1+1+2+3 Secondly, the three cases are regarded as three different types of calculations.

The first category: there is a box with 4 balls, and the rest of the boxes put 1 ball each, because the 4 balls can be placed in one of the different four boxes, there are 4 ways to put it, and the other boxes only put one ball, and the balls are the same, any exchange is the same way to put it, so the first type only has 4 ways to put it. Category II:

There is a box with 1 ball, there are 4 ways to put it, and the rest of the boxes are put with 2 balls, which is the same as the first type, and any exchange is the same way, so the second type also has only 4 ways to put it. Category 3: There are two boxes with one ball each, and the other two boxes with 2 and 3 balls respectively

The first step is to take two of the four boxes and put a ball each, which has 24 methods. The second step is to put 2 balls and 3 balls into the remaining two boxes, which can be placed in P22 ways. According to the principle of multiplication, there are c24 p22 = 12 ways to put it.

According to the principle of addition, there are 4+4+12=20 different ways to meet the requirements of the question. Answer: There are 20 different ways to put it. This problem can also be seen as putting a ball at the bottom of each box first, so that the box is not empty, leaving 3 balls, and putting in 4 different boxes.

There are 4 ways to put the first ball. There are also 4 ways to put the second ball,......There are 4 ways to put the seventh ball. According to the principle of multiplication, there are a total of 4 4 4 4 4 4 4 4 = 16384 (types).

Summary. There are 125 possibilities for 5 boxes with 3 balls.

There are 125 possibilities for 5 boxes with 3 balls.

I'm still a little confused, can you be more detailed?

Let's say we have 3 balls and 5 boxes, it is easy to think of placing three balls into the box manually so that we can get the result by counting visually. However, if the number of balls and boxes is large, the pose-empty state method is not suitable, and we need to consider using a combination method to solve the problem. One way is to adopt the idea of permutations and combinations, the trace source sees the balls as small balls of different colors, so in the case of 5 boxes, each ball has 5 placement options, then according to the principle of multiplication, the total placement of 3 balls may be 5 times 5 times 5 times 5 equals 125.

So, there are a total of 125 different ways to place it, and it's up to us to answer the question.

If each box does not need to be full.

Then there are 5 ways to put 1 ball.

So there are 5*5*5*5*5=5 5=5 5=3125 kinds.

If every Chi Brother box has to have a bent Danzhou ball.

So there are 5 types of the first ball, 4 types of the second ball, 3 types of buried for the third ball, and 2 types of the fourth ball.

There is 1 type of the fifth ball.

So there are 5*4*3*2*1=120 kinds.

1.Different balls have different boxes, and for balls, each ball has 3 choices, so it is 3 to the fifth power.

2.For the same ball and the same box, there are 500, 410, 320, 311, 221 by enumeration.

3.Different balls with the same box stove Minzi, according to the 5 cases of the above code argument surface to find the combination and then sum, there are 1 + 5 + 10 + 25 + 30 = 71 kinds.

4.Different boxes for the same ball, there are 3+6+6+3+3=21 types.