Arrangement and combination of the problem, the mathematical Emperor help 1. There are ten identical books and three different boxes. At least one book is put in each box. How many ways are there?

C(2,9)
It is equivalent to the baffle problem. There are nine gaps in the middle of ten books. If two baffles are inserted into them, they are divided into three parts. There are 36 methods in total

At present, 5 students from a, B, C, D and E are arranged to participate in a volunteer service activity. Each student is engaged in one of the four jobs of translation, tour guide, etiquette and driver. At least one person participates in each job. A and B can't drive but can engage in the other three jobs. C, D and E can all win the four jobs. Then the total number of different arrangements is______ (fill in figures)

According to the meaning of the topic, the discussion is divided into two situations: (1) Party A and Party B participate in one of the three jobs except driving: C31 × A33 = 18; (2) Party A and Party B do not participate in one job at the same time, and then they are divided into two small situations; (1) two of the three people (1 ° C, D, Xu) undertake the same job, and there are A32 × C32 × A22 = 3 × 2 × 3 × 2 = 36; (2) Party A or Party B and one of the three people (C, D, Xu) undertake the same job: a 32 × C31 × C21 × A22 = 72 species; according to the principle of classification and counting, there are 18 + 36 + 72 = 126 species, so the answer is: 126

Divide a set of 54 playing cards equally to 6 players and try to find the probability of double king and 4 A's in one player's hand
If you are the one
Neither of the answers given is correct
If you have the ability to answer, please continue

The denominator is C54 9 C45 9 C36 9 C27 9 C18 9
Molecule 6 × C48 3 C45 9 C36 9 C27 9 C18 9

C16 x²+3x+2 =C16 5x+5

The direct result is X & sup2; + 3x + 2 = 5x + 5, and the solution is x = - 1 or x = 3
That's all
Pay attention to the adoption

Seeking the formula for calculating the probability of each prize in lucky lottery 30-7 is the application of number permutation and combination in mathematics
I don't want a special number. I just need a positive number. For example, how to calculate the probability of 6 middle numbers and how big it is, and how to calculate the probability of 5 middle numbers. The total middle number should be C (30,7). But if 6 middle numbers are obviously not c (30,6), how to calculate it? I just need the calculation formula of 6 middle numbers and 5 middle numbers, I've returned the math Scripture to my math teacher. I've forgotten. I'm good at math, lottery fans,

Choose 7 from 30, and calculate the probability of 6 signs as follows: C (7,6) times C (23,1) divided by C (30,7)
Choose 7 from 30, and calculate the probability of 5 numbers as follows: C (7,5) times C (23,2) divided by C (30,7)
I'm a math teacher in high school. There should be no problem

N small balls with labels 1 to n are respectively placed in boxes with numbers 1 to N, one for each box. It is required that the number of small balls should not be the same as that of the box. How many kinds of points are there?
How to ask? Please explain what you ask for

This is a famous envelope problem. Many famous mathematicians have studied it. Euler, a Swiss mathematician, gives a recurrence formula according to the general situation: using a, B, C An envelope with the names of n friends, a, B, C The total number of misplaced letters is f (n). Suppose that a is misplaced in B

(1) Conjecture C (0, n) + C (1, n) + C (2, n) + +C (n-1, n) + C (n, n)
(2) Can we use the previous question to find the number of subsets of a set?
(the numbers in brackets are superscripts on the left and subscripts on the right)

(1) It can be concluded from binomial expansion
（1+1）^n=C（0,n)+C(1,n)+C(2,n)+…… +C(n-1,n)+C(n,n)
So C (0, n) + C (1, n) + C (2, n) + +C(n-1,n)+C(n,n)=2^n;
(2) For a set of n elements, the number of elements in its subset may be 0, 1, 2 ,N
The number of subsets C (0, n) containing only 0 elements;
The number of subsets C (1, n) with only one element;
The number of subsets C (2, n) with only two elements;
……
The number of subsets containing n elements is C (n, n);
So the number of subsets is 2 ^ n

(1) How many different allocation methods are there when 10 excellent quota are allocated to 6 classes with at least one in each class?
One more question
How many allocation methods are there for each school with at least one quota and different quotas?

Baffle method:
Nine of the 10 places are vacant,
Divide the 10 places into six, at least one for each,
Then we just need to put a partition on each of the five gears in the nine gears, so that the 10 indexes can be divided into six parts from left to right, and each part can meet at least one quota. We can solve the problem by giving the six parts from left to right to class 1, 2, 3, 4, 5 and 6 in turn
The number of different methods of putting 5 partitions on 9 neutral gears corresponds to the number of quota allocation methods that meet the requirements
5C9＝126.
Or enumeration
1 + 1 + 1 + 1 + 1 + 5 A6, 6 species
1 + 1 + 1 + 1 + 2 + 4 A6 take 230 species
1 + 1 + 1 + 1 + 3 + 3 C6 take 2 15 species
1 + 1 + 1 + 2 + 2 + 3 A6 3 divided by 2 60
1 + 1 + 2 + 2 + 2 C6 take 4 15 kinds
There are 126

(2010. Tangshan three mode) 7 cards were written on figures of 1, 1, 2, 2, 3, 4, 5, respectively.
A. 198B. 156C. 145D. 142

The mantissa is 1. Among the remaining 6 numbers, 2 and the other 4 numbers are selected at the same time to form 4 * 3 = 12 kinds of three digits. Only one 2 is selected, A53 = 60. To sum up, 60 + 12 = 72 and the mantissa is 3, 5. In the same discussion, take 3 at the end as an example. 2 and 2 are selected at the same time, 3 × 3 = 9 with the other 3 numbers

Three boys and three girls stand in a row. If girl a doesn't stand at both ends and only two boys are adjacent to each other, the number of different arrangements is ()
A. 360B. 288C. 216D. 96

First of all, there are 432 kinds of c32a22a42a33 in the three boys with only two adjacent permutations, 144 kinds of 2 × c32a22a32a22 in the three boys with only two adjacent permutations, and 288 kinds of different permutations

Arrangement and combination of the problem, the mathematical Emperor help 1. There are ten identical books and three different boxes. At least one book is put in each box. How many ways are there?