A mathematical problem of permutation and combination Party A and Party B go to the park together. They independently choose 4 scenic spots from No. 1 to No. 6 for sightseeing. Each scenic spot is visited for 1 hour. What is the probability that they will be in the same scenic spot in the last hour?

A mathematical problem of permutation and combination Party A and Party B go to the park together. They independently choose 4 scenic spots from No. 1 to No. 6 for sightseeing. Each scenic spot is visited for 1 hour. What is the probability that they will be in the same scenic spot in the last hour?

Choose the last scenic spot first: there are six choices
There are three left: 5 * 4 * 3, two people, so there are still * 5 * 4 * 3
If you choose any four scenic spots from No.1 to No.6 for sightseeing, the general choice is: 6 * 5 * 4 * 3 for two people, so you have to * 6 * 5 * 4 * 3
The probability is 6 * 5 * 4 * 3 * 5 * 4 * 3 / 6 * 5 * 4 * 3 * 6 * 5 * 4 * 3 = 1 / 6
So the probability is 1 / 6

A mathematical problem of permutation and combination
If there are 10 pieces of the same kind, you can take at least one at a time, or you can take all of them at one time without putting them back. How many different ways are there?

It is equivalent to putting several separators between 10 flags
Take 10 times to get 9 separators
Nine times is eight separator
An analogy
Therefore, there are 512 kinds of C (9,9) + C (9,8) + C (9,7) +. + C (9,1) + C (9,0) = 2 ^ 9

A mathematical permutation problem
There are 10 people who can row in a small boat. There are 3 people who can only row in the left paddle, 4 people who can only row in the right paddle, and 3 people who can row on both sides. Six of them are selected to share the two paddles in the small boat. Regardless of the order of the three people on the same side, how many choices are there

1. The left side is composed of three people who can only paddle left, so the right side is composed of the other seven people, C7 (4) = 352, and the left side is composed of two people who can only paddle left, so we need to draw one from the three people in the metropolis, so the combination = C3 (1) * C3 (2) = 9, and the right side is composed of four right oars and the remaining two people who can paddle left, the situation = C6 (3) = 2

A mathematical problem (permutation and combination)
A=1,2,3,4,5 B=6,7,8
These two sets
Make up a 6-digit number
It is required to select 2 different elements from a
Choose four elements from B, the same or different
How many situations are there

There are 24300 species. There are 10 species for 2 in a and 4 for 3 elements in B
1. There are only one element in B, and there are three ways to choose. However, there are 5 * 6 = 30 ways to insert the four same numbers in a, so there are 5 * 6 * 3 * 10 = 900 ways to choose
2. There are two elements in B, and there are three combinations. For example, there are three combinations of 6 and 7, 6777666667. Therefore, there are three * 3 = 9. There are six arrangements like 6677 (three at the beginning of 6, three at the beginning of 7), and there are five * 6 = 30 in a, so there are six * 30 = 180, and so are 6688 and 7788, so there are 180 * 3 = 540, 540 * 10 = 5400 (10 refers to the selection of 2 for type A). There are also numbers similar to 6777, with 4 permutations and 5 * 6 = 30 for a to insert, so there are 4 * 30 = 120, while there are 6 numbers similar to 6777, so 120 * 6 * 10 = 7200
3. In B, there are three combinations of three elements (one of the three elements is always taken twice). For example, 6678, you can use 7 and 8 to insert two 6, there are 3 * 4 = 12 permutations, and then 12 * 5 * 6 = 360, and there are three similar 6678, so 360 * 3 * 10 = 10800
To sum up, there are a total of 900 + 5400 + 7200 + 10800 = 24300 species
My senior three, typing is not easy

On a problem of permutation and combination in mathematics
Given the five numbers of 12345, how many different numbers are there

The number of reusable numbers is: 5 * 5 * 5 = 125,
If each number can only be used once, it is: 5 * 4 * 3 = 60

At present, six scenic spots are allocated to six tourists, including two tickets for scenic spots a and B, and one ticket for scenic spots C and d. There are different allocation methods______ Species (answer with numbers)

According to the meaning of the question, this is a step-by-step counting principle, which can be regarded as dividing six different customers into four scenic spots. A and B each have two places, C and D each have one place, and the number of allocated species is c62c42a22 = 180, so the answer is: 180

Senior High School Mathematics Series derivation formula encyclopedia, the most primitive formula
What I want is the formula of the sequence, not the whole high school.

The accessory is easy to use!

The solution set of inequality x ^ 2-x-2 is ()
2, inequality x ^ 2 + X + 1 | 5x-3 | - | 4x + 1 |, the solution set greater than 0 is ()
3. The solution set of inequality 5x-1 + radical (2x-1) > 4x-2 + radical (2x-1) is ()
4, the solution set under | radical (1-1 / x) minus 2 | 3 is ()
The solution set of inequality x ^ 2 + X + 1 (| 5x-3 | - | 4x + 1 |) greater than 0 is ().

1、
-1≤x

1. If the image of the quadratic function y = f (x) passes through the origin, and 1 is less than or equal to f (- 1) less than or equal to 2, 3 is less than or equal to f (1) less than or equal to 4, then the value range of F (- 2)_______________
2. M = log subscript a (1 + a), n = log subscript a (1 + 1 / a), then M_____ N (compare size)

First question
Let f (x) = ax ^ 2 + BX
From 1 to f (- 1) to 2,3 to f (1) to 4
(ba-1 generation)
Two inequalities are obtained
1 less than (a-b) less than or equal to 2 let this be inequality 1
3 less than or equal to (a + b) less than or equal to 4 let this be inequality 2
Multiply inequality 1 by 3 and add inequality 2
The inequality 6 less than (4a-2b) less than or equal to 10 is obtained
And f (- 2) equals (4a-2b)
Therefore, the value range of F (- 2) is greater than 6 and less than 10
The second question (this text narration is relatively difficult, let me give a general description)
Used as the difference method
M-N is reduced to 1 (where a is greater than - 1)
So m is greater than n

1.x^2+2x-15>0
2.x^2>2x-1
3.x^2

1.x^2+2x-15>0
（x－3)(x＋5)＞0
X ＞ 3 or X ＜ - 5
2.x^2>2x-1
x²－2x＋1＞0
(x－1)²＞0
x＞1
3.x^2