High School Permutations 5, High School Permutations?

It is equivalent to 5x+3y=13, x can take 1,2, y takes 3,1, the title is two 5s multiplied by random numbers, so it can be regarded as 5 (m+n), where x=m+n, and here there are (0,1)(0,2)(1,1)(1,0) and (2,0), so there are five ways to take it, and there are 1000 probabilities for the total event.

5x 5y 3z 13, x, y, z each have ten numbers to be taken, so there are 10x10x10 1000 combinations, 5 times any number to get the result you will only be 0 or 5, the single digit of the result is required to be 3, then z can only be 1,1000 combinations of the results that can make the equation stand (1,1,1), (2,0,1), (0,2,1), so the probability is.

Is high school so difficult, I went to college in vain,

<> I calculated that the answer is 31 60, but it is not guaranteed to be right.

First choose the mandatory type of Sullen Tomb 1

Then choose two C2 3 from the remaining 3 types of the state hood

There are three cases of A3 in the exchange order in the land register.

So, 1*c2 3*a3 3=18.

5 times of 3 plus 4 times of 3 plus 3 times of 3 plus 2 times of 3 = 360

1) From the permutations of the four different elements a, b, c, and d, take out the permutations of the three different elements as? 4！=24

2) How many different arrangements are there for 4 boys and 4 girls in a row, and girls are not in a row at both ends? a(64)*a(44)=8640

3) 3 RMB with 1 jiao, 1 RMB with 5 jiao, and 4 RMB with 1 yuan, how many different currencies can be formed with these RMB? 4*2*5-1=39

4) Use the six numbers 0, 1, 2, 3, 4, and 5 to form a five-digit number with no repeating numbers.

How many odd numbers are divisible by 5 Divisible by 15 Smaller than 35142 Smaller than 50000 and not a multiple of 5 3 4a(43) =288 4a(43)+a(54)=216 2a(44)+3a(33)=66 2a(54)+4a(43)+2a(22)+1=347 a(42)a(44)=288

5) 7 people in a row, how many different arrangements are there in each of the following situations?

A row head a(66) = 720

A does not row the head, nor does it row the tail 5a (66).

A, B, and C must be together a(55)a(33)=720

There are only two people between A and Ba[52]a[44]a[22]=960

A, B, and C are not adjacent to each other, a[44]a[53]=1440

A is to the left of B (not necessarily adjacent).

A, B, and C are in the order from high to low, to left to right, a[74]=840

A does not rank first, and B does not rank a[77]-2a[66]+a[55]=3720

6) Take any 3 of the 5 numbers 2, 3, 4, 7, and 9 to form a three-digit number without repeating numbers.

How many such three-digit numbers are there? a[53]=60

What is the sum of the digits in the single digits of all these three-digit fronts? (2+3+4+7+9)a[53]/5=300

The sum of all these three-digit numbers is 300*[100+10+1]=33300

8. Individual numbers 1-8

Let's say you pick 1-4 in circle A, and the rest are in circle B; And if you select 5-8 in A and 1-4 in B, these two situations are the same, because there is no difference between the two circles of A and B. 4 out of 8 people is c84... But there is no difference between the two circles, so divide by a22

Choose four people to form a circle of C84, there is no difference between the two circles, so except for A22.

For example, if you choose a group of 1, 2, 3, and 4, it is the same as if you chose a group of 5, 6, 7, and 8.

The first division: there is a group of 3 people, the other two groups of 1 person each, a total of C35 A33 = 60 different ways to participate, the second division: there is a group of 1 person, the other two groups of 2 people, a total of C25 C23 A33 = 90 different ways to participate.

There are 60 + 90 = 150 ways to participate.

Then calculate the participation method of A and B to participate in the same project, and regard A and B as 1 person, a total of 4 people, and the possible participation methods are: C24 A33 = 36 The number of arrangement methods that meet the above requirements and A and B do not participate in the same project is 150 36 = 114

What I learned in high school is similar to what is explained in the encyclopedia, so it should be understandable. There are many solutions to a topic, and the expression of different processes will be different, but it does not mean that it is incorrect, there is no standard answer to this kind of question, but as long as the idea is correct, it is not far from success.

Concept: Permutations and combinations are the most basic concepts of combinatorics. The so-called arrangement refers to taking a specified number of elements from a given number of elements and sorting them.

Combination, on the other hand, refers to taking only a specified number of elements from a given number of elements, regardless of sorting. The central issue of permutations and combinations is to study the total number of possible scenarios for a given required permutations and combinations. Permutations and combinations are closely related to classical probability theory (classical generalizations in high school).

Definitions and Formulas:

The definition of permutation and its calculation formula: from n different elements, any element of m (m n, m and n are natural numbers, the same below) are arranged in a certain order, which is called taking out a permutation 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). a(n,m)=n(n-1)(n-2)……n-m+1)= n!

n-m)!In addition, it is stipulated that 0!Definition of =1 combination and its formula:

Taking any m (m n) elements from n different elements and forming them into a group is called taking 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). c(n,m)==a(n,m)/m！

c(n,m)=c(n,n-m）。(n>=m) Other Permutations and Combinations Formula The number of cyclic permutations of m elements taken out of n elements = a(n,m) m=n!/m(n-m)!.

N elements are divided into k classes, and the number of each class is n1, n2, ,..The full permutation of nk is n!/(n1！

n2！×.nk!

K class elements, the number of each class is unlimited, and the number of combinations of m elements taken from it is c(m+k-1,m).

The number of permutations, take m from n and arrange them down, there are n(n-1)(n-2).n-m+1), i.e., n!/(n-m)!

The number of combinations, take m from n, which is equivalent to not arranged, that is, n!/[(n-m)!m!]

1. There are special circumstances in the permutation and combination must be given priority, such as the single digit in this problem is 6, and the thousand digit cannot be 0 should be ranked first, your second algorithm is correct, the first one is wrong, why is it wrong? For example, when the ten digits are 0, there are 8 options for the 100 digits, and 7 options for the 1,000 digits instead of 6! Antan Dan.

2. If the question specifies that different crops must be planted on each piece of land, there are 18 methods of c(3,2)*a(3,3); Otherwise, 4*4*4-3*3*3=37 methods.

Your algorithm doesn't take this situation into account, and the fake Lu Xian loses like ten sedan positions and hundreds of places don't get 0, and thousands of places get 0

So if you want to reduce the situation early, the result will be the same.

Analyze the problem, choose 3 out of 4, including A, so you can only choose 2 out of 3, arrangement: c(3,2)*a(3,3)=18

The format of the first and fourth digits should be 6, then the first digit has only 8 possibilities except 0 and 6, the second digit has 8 possibilities except for the first and last 6, and the third digit has 7 possibilities after removing the bits, so the total should be changed to 8*8*7 = 448

Second, first choose one seed from the three fields C31, and then choose two from the other three seeds, because the land is different, so it is necessary to arrange it, that is, A32, so the answer should be C31*A32 = 3*6=18

The number can be repeated in imitation of 3700 mu, as long as the first one is not 0.

18 pairs of Bishan's Because the other two are selected first and carefully seeded c32, and then all are arranged a33, the answer is 18