Permutation and combination high school mathematics "Six people, each of them has a hat, but each of them is required to wear someone else's hat. How many kinds of wearing methods are there?" I would like to ask enthusiastic netizens how to solve this type of problem?

This is a dislocation problem. Remember that the general term formula an an = (n-1) (a (n-1) + a (n-2)) A1 = 0 A2 = 1 A3 = 2 A4 = 9 A5 = 44 A6 = 265
This is a classic problem in permutation and combination

How many ways can five interns be assigned to three classes of senior one, with at least one teacher in each class and two at most? I did it in two ways: The first is: C (5,1) C (4,2) +C (5,2) C (3,1) +C (5,2) C (3,2) =90 The second is: C (5,2) a (3,1) C (3,2) a (2,1), two teachers from five classes, one from three classes, two teachers from the remaining three teachers and one from the remaining two classes But the result is 180 What's wrong with the second solution?

Because the second method is repeated, assuming that five teachers are ABCDE and three classes are 123, then consider these two situations: case 1: the first step selects teacher AB, class 1, the second step selects teacher CD, class 2: the first step selects teacher CD, class 2 selects teacher AB, class 1 These two situations are actually the same

Roll dice twice, known points are different, find the probability of at least one 6 points

For a long time, I haven't done such a question for a long time. I'll answer, nonsense, no responsibility for right or wrong. First of all, the probability of six points in a single roll is one sixth. Then the probability of six points in two rolls should be the sum of the single probability, that is, one third

In the process of information transmission, an arrangement of four numbers (the number can be repeated) is used to represent a message, and different permutations represent different information. If the numbers used are only 0 and 1, then there are at most two numbers corresponding to the information 0110, and the number of the same information is () A. 10 B. 11 C. 12 D. 15

There are at most two numbers in the corresponding position with the information 0110. The same information includes three types: the first one is C42 = 6 (numbers) which are the same as the numbers in two corresponding positions of information 0110. The second category is that C41 = 4 numbers are the same as the number in one corresponding position of information 0110

In a numerical matrix with n * n numbers, the top row has n values that are different from each other. Can these n values form the remaining rows in different order and make the order of any two rows different? If there is a matrix with M rows and each row has n different values, in order to make each row not repeat, What's the value of M? Isn't the third sentence and the fourth sentence the same meaning? Please say the train of thought and the specific meaning of the title before you answer. I can't understand the meaning of the title directly

The answer to the first question is yes
For the second problem, which is to calculate the ordered arrangement of N numbers, the maximum value of M is an n (superscript) n (subscript)

The factorial of one plus the factorial of two is added to the factorial of N. how much is this sum divided by the factorial of N?

Lim n - > infinity 1! + 2! + 3! + n! / N! = 1 + 1 / N + 1 / [n (n-1)] + 1 / [n (n-1) (n-2)]... + 1 / N! = 1

Calculate 1! +2！ +3！ +… +100！ The digits of the number obtained are______ .

Because of 5! ，6！，… ，100！ It's 2 × 5,
Then starting from 5, all the bits of the factorial are 0,
Just look at 1! +2！ +3！ +4！ Only one bit is enough
By 1! +2！ +3！ +4！ =33，
So calculate 1! +2！ +3！ +… +100！ The digit number of the obtained number is 3
So the answer is 3

We don't need to define the n-factorial. For example, we can get 120 by substituting n = 5 It cannot be emphasized again and again by definition that the row form of a fraction is the same as the first n terms and formulas

Programming with an algorithm will not come out

What is the relationship between the last digit of the factorial result of a number and it is best to give a function expression

0! = 1,
1! = 1,
2! = 2,
3! = 6,
4! = 24, last 4
Factorials above 5 must be multiples of 10 because they contain both 5 and 2. The last digit is 0

1+1+1/2!+1/3!+.+1/n!

How high are you? If you are 3 years old, one is very simple
Let an = 1 + 1 / 2! + 1 / 3! +. + 1 / N!
Bn＝1+1/2+(1/2^2)+(1/2^3)+...+(1/2^n)
Suppose an < BN holds,
Then an + 1 = an + 1 / (n + 1)!
Bn+1=Bn +1/2^(n+1)
And (n + 1)! > 2 ^ (n + 1) > 0 [(n + 1)! = 2 * 3 * 4 * 5... * (n + 1) > 2 * 2 * 2... (n + 1) 2]
So 1 / (n + 1)! < 1 / 2 ^ (n + 1)
The + 1 + 1 + 1 + 2 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 + 1 + 1 / 2 + 1 + 1 / 2 + 1 + 1 / 2 + 1 + 1 / 2 + 1 + 1 / 2 + 1 + 1 / 2 + 1 + 1 / 2 + 1 + 1 / 2 + 1 + 1 / 2 + 1 + 1 / 2 + 1 + 1 / 2 + 1 + 1 / 2 + 1 + 1 / 2 + 1 + 1 /

Permutation and combination high school mathematics "Six people, each of them has a hat, but each of them is required to wear someone else's hat. How many kinds of wearing methods are there?" I would like to ask enthusiastic netizens how to solve this type of problem?