Permutation and combination of mathematical probability problems There are five different toys for all three children, and the probability of at least one toy for each child is calculated

Permutation and combination of mathematical probability problems There are five different toys for all three children, and the probability of at least one toy for each child is calculated

The probability of giving at most two people = C (3,2) * (2 / 3) ^ 5-c (3,1) * (1 / 3) ^ 5 = 32 / 81-1 / 81 = 31 / 81
So the probability of giving at least three people is 1-31 / 81 = 50 / 81

A super simple mathematical problem of permutation and combination probability in Senior High School
There are the following maps in the city: (the black line is the road, please ignore the place where the section is broken, and the special node is marked as e point, which has four roads horizontally and four roads vertically)
┏┳┳┓
┣╋╋┫
┣┼╋┫
┗┻┻┛
Starting from the lower left corner, you can only walk right or upward, and the probability of making a choice is equal, until you reach the upper right corner, ask the probability of not passing through point E
A.1/2 B.2/5 C.3/5 D.2/3
A or B,
The reason of choosing a: every turning point has the same probability of going up and right, which is 1 / 2, so the probability of reaching e is 1 / 2.
The reason for choosing B: there are 20 ways of C (3,6), and there are 8 ways of C (1,4) after "digging out" point E, so it is 2 / 5.
So, which one?

1/2,
Whether or not to go through point E is completely determined by the first two steps
Top right
Right, right
Up right
Up
Two of them cross the e-point: upper right, upper right
Because the probability of making a choice is equal, so the probability of the above four cases is 1 / 4, and the probability is 1 / 2
Add: first of all, emphasize the above sentence:
Whether or not to go through point E is entirely determined by the first two steps
B in your supplement implies that the probability of each path is the same. But in fact, the probability of the path that reaches the top first or the right first is higher. For example, the probability of the upper right on the path is 1 / 2 × 1 / 2 × 1 / 2 × 1 × 1, because there is only one way to go after reaching the edge. The probability of the upper right on the path is 1 / 2 × 1 / 2 × 1 / 2 × 1 / 2 × 1 / 2 × 1
So B is wrong

How to distinguish permutation and combination in Mathematics

The common point of permutation and combination is to take any m (m ≤ n) elements from n different elements. The difference is that permutation is arranged in a certain order, and combination is combined into a group regardless of the order. Therefore, "order" and "disorder" are important signs to distinguish permutation and combination
There are 11 senior one students in the Union: 1. How many letters did each two exchange? 2. How many times did each two shake hands?
2、 There are 10 students in the extracurricular activities group of senior one: 1. How many different ways to choose a leader and a deputy leader? 2. How many different ways to choose two students to participate in the mathematics competition?
1、 1. Because every two people exchange a letter, the letter from a to B and the letter from B to a are two different letters, so it is related to the order
2. Because every two people shake hands once, a and B shake hands, B and a shake hands is the same shake hands, which has nothing to do with the order, so it is a combination problem
1、 1. It's a permutation problem, common = 110
2. It's a combinatorial problem, with a total of 55 handshakes
2、 1. It's a permutation problem, with a total of = 10 × 9 = 90
2. It is a combinatorial problem, with 45 kinds in total

How to distinguish a and C in mathematical permutation and combination?

Classmate, this problem focuses on understanding
A refers to the arrangement. Arrangement is like queuing. Objects are ordered
C refers to combination. Combination is like fried rice with eggs and fried eggs with rice
Because of their different meanings, the calculation method is close to the following
A（x,y）=y!/（y-x）!
C（x,y）=y!/【（y-x）!*x!】
Where y > = X
Deep understanding of concepts is a good way to solve science problems logically. What is deep? It depends on your own understanding

A master of mathematical permutation and combination probability
In a box, there are 15 balls, one red, one yellow, one blue, one green, and 11 black balls. Take one out, put it back and continue to touch. The probability of each ball being touched is the same
Touch 42 times, in 42 times, can touch at least one red one yellow one blue one green probability is how many?

Because every time the probability of red, yellow, blue and green is the same, it is 1 / 15
In 42 times, the probability of the ball not appearing at least some color ball is (14 / 15) ^ 42,
In 42 times, the probability that at least two kinds of color balls didn't appear in the ball was (13 / 15) ^ 42
In 42 times, the probability that at least three kinds of color balls didn't appear in the ball was (12 / 15) ^ 42,
In 42 times, the probability of no color ball is (11 / 15) ^ 42,
According to the principle of capacity and disassembly, the probability is
1-C(4,1)×(14/15)^42+C(4,2)×(13/15)^42-2×C(4,3)×(12/15)^42+6×(11/15)^42=
79.35%

I don't know about permutations
There are three cards marked with letter A and six cards marked with numbers 1 to 6. If any six of them are used to form a brand, what is the total number of different brands

The classification of three alphabetic cards is discussed
C (6,3) * a (4,3) = 144
Choose two: a (6,4) * C (5,2) = 3600
Choose one: a (6,5) * C (6,1) = 4320
Not selected: a (6,6) = 720
Add up to 8784

Permutation and combination exercises
There are five different books for four people. How many ways are there?
How many ways to divide five identical books among four people?

If each person has at least one book (1) choose two of the five books for one person, and the remaining three books for the remaining three people, there are a total of C (5,2) * C (4,1) * P (3,3) = 240 kinds (2) one person gets two books, and there are four ways to divide books. (1) 4 * 4 * 4 * 4 = 1024 kinds (2) partition method_ There are six vacancies

There are five different books, including two Chinese books, two mathematics books and one physics book. If they are randomly placed on the same shelf, the probability that the books of the same subject are not adjacent is ()
A. 15B. 25C. 35D. 45

According to the meaning of the topic, we know that the topic is the probability of an equal possible event. The event involved in the experiment is to randomly put five books on a shelf, with a total of A55 = 120 results. We study the number of non adjacent permutations of the same kind of books by classification. Assuming that the first book is a Chinese book (or a mathematics book), and the second book is a mathematics book (or a Chinese Book), there are 4 × 2 × 2 × 1 = 32 possibilities; assuming that the first book is a Chinese book If the second book is a physics book, there are 4 × 1 × 2 × 1 × 1 = 8 possibilities. If the first book is a physics book, there are 1 × 4 × 2 × 1 × 1 = 8 possibilities. The probability that the books of the same subject are not adjacent is p = 48120 = 25, so choose B

A mathematical problem about permutation and combination
How many quadruple odd numbers larger than 2000 can be formed with 12345

C1 / 2 first even last odd C1 / 3 middle two A3 / 3 or C1 / 2 first odd last odd C1 / 2 middle two A3 / 3. Add these together!

A mathematical problem of permutation and combination
There are five floors in the teaching building. Each floor has two stairs. The ways from one floor to five floors are ()
a 2^5
b 2^4
What? Why?

B, because there are four stairs in five floors