Sophomore Mathematics Permutations and Combinations Knowledge Points Kneel and beg!!!!!!!!

Step-by-step calculations. The first is grouping. c(6,44*c(2,1)*c(1,1)=6!/4!/(6-4)!*2*1=30。

In the second step, three different groups are sorted, two of which are identical, for a total of 3!/2!=3。

The two are superimposed, 3 * 30 = 90 schemes.

Enumeration verification can be programmed and the results are correct.

Attached: Enumeration results and fortran**.

Tried the math, hope it helps.

Here's how, please refer to:

Arrange the problem. There are 1, 2, 3, 4, 5, 6, 7. There are 3 even numbers and 4 odd numbers.

c23, (2 even numbers out of 3 even numbers) c34 (3 odd numbers out of 4 odd numbers) a55 (for full permutation). c23*c34*a55

2.Five digits, even numbers are on even digits, so 1,3,5 digits are odd numbers, and 2,4 digits are even numbers.

c23, (2 even numbers out of 3 even numbers), c34 (3 odd numbers out of 4 odd numbers), a33 (all 3 odd numbers picked out), a22 (all out of 3 even numbers).

C23*C34*A33*A22 Look!

The first number in parentheses is in the upper right corner of c, and the second number is in the lower right corner of c.

4）*c(2,4)+c(1,2)*c(3,4)*c(3,4)+1=105

There is a question that 4 can only speak English, 4 can only French, 2 can do anything, and the selected group must have 4 people who can speak English and 4 who can speak French, and the number of people is not limited.

Assume 3 scenarios.

If 2 people who know everything go there, 2 of the 4 will go to the French, and 2 will go to the English.

If you go to 1, you must go to 3 of the 4 who can only speak French, and 3 who can only go English.

If they didn't go, 4 of them could only speak English and 4 of them could only go in French.

2.Belongs to the category that has no talent and doesn't know how to learn.

3.How to summarize it for you on the computer, you can only read it when you write it on paper, or solve it in person, you can ask a tutor for this.

4.I can't point on the computer, for the same reason.

There are 10 members, six of whom speak English, six of whom speak French, two of whom speak both English and French, and four of the remaining eight who speak English and four of whom speak French. That is, choose 2 people who can both English and French from 10 people, that is, C10 2Another 4 out of 8 people who can speak British are C8 4, and the next 4 people can only speak French.

The situation is c10 2 times c8 4

C10 2 45, C8 4 70, 45 * 70 3150 cases.

Your problem is very easy to think of, permutations and combinations are relatively simple in mathematics, but not very useful in life.

It's simple, but it's not so good that you can say a few words in front of the computer. How to learn well, love can't help.

1.Since there are only 10 members, 6 of whom can speak English and 6 of whom can speak French, it means that 2 of them can speak French and English, and the rest are a team of 4 people who can only speak French and 4 people who can only speak English. So there are three forms of selected groups:

1) There are no two of them in the group: 4 of them are required to be able to speak English and 4 of them can speak French, then you have to choose all the rest, which is a situation.

2) There is 1 team in the group: that is, one of the two will be selected, and then a team of only 3 people will be selected from the other teams that can only speak French and English, that is, 2*2*4=16

3) There are 2 people in the group: then choose 2 people from each of the teams that can only speak French and English, 4*3*4*3=144

The total number of cases obtained: 1 + 16 + 144 = 161

I don't know if it's right, I'm a senior in college, and I almost forgot about it!

2.I was like that when I was in high school, it was just a practice feeling, hehe!

The mobile phone answers only to answer the question There are two people who will The first is not to choose the two 1 kind 2 choose one So you have to choose three C2 1 C4 3 C4 3 = 32 All three are selected Then just choose two more C4 2 C4 2 = 36 for a total of 69

ps It seems that the recommendation is wrong, and the first item should not be multiplied by 2.

Purpose: Choose 8 out of 10 people who can speak at least one of English or French.

1. If 6 people can speak both English and French. There are only 6 people who can speak English or French, excluding 2, if 5 people can speak both English and French. There are only 7 people, excluding people who can speak both English and French, exactly 8 people.

2 English + 4 English and French + 2 French There are 9 people who can speak both English and French. 3 English + 3 English and French + 3 French There are 10 people who can speak both English and French. 4 English + 2 English and French + 4 French A person who can speak both English and French needs 11 people, but there are 10 people in total, so it doesn't exist.

Under this analysis,It's out.,Maybe my analysis is more complicated.。。

This kind of must be drawn, English: . .

French:. The middle 2 people will, up and down are accounted for, 3 kinds of situation analysis, the middle 2 people went, went to one, didn't go, it's very easy to calculate, c, not a, no need to arrange, graduated from college, almost forgotten.

Learn according to the type of question, for example, you are talking about generalist questions. Categorize by all-rounder. The idea of categorical discussion transforms uncertain problems into deterministic problem solving.

I've done this before! But now that I'm graduating from my senior year, I've forgotten all the previous scoring questions. Ay.

I'll tell you little by little.

a(7,2).

The total number of cases in which two of the seven are chosen at any time du,a(zhi3,1)、a(10,3) is the same,First of all, 3 represents 3 people,Since there is one person who wins the lottery,Then three people have 3 possibilities,So multiply three,10 lottery tickets contain 3 winning lottery tickets,There are 7 non-winning lottery tickets,a(7,2) is the situation that two of the 3 people get the winning lottery tickets,a(3,1) is the situation that the remaining one person gets the winning lottery ticket, Multiply to get the total number of cases where a person gets the lottery ticket, so it seems that it is also necessary to multiply a 3 to represent any of the three people, and the total number of cases of three people is a(10,3), divide it to get the probability.

I hope it can help you, if you don't understand, you can ask.

For everyone, the probability of winning the lottery is 3 out of 10

There are three scenarios in which exactly one of the first 3 people wins the lottery.

They are: the first person to win the lottery, the first.

2. Three people do not belong to the middle;

The second person wins, the first.

1. Three people missed;

The third person wins, the first.

One or two people missed.

The probability of the above three cases is (3 10) * (7 10) * (7 10) = , so the probability of one person winning the lottery is.

Multiply by 3 because it is the one of the first three people who wins the lottery, that is, it can be the first, the second, and the third.

a(7,2)*a(3,1) is a combination of two randomly drawn from seven unwinning tickets and multiplied by one of the three winning tickets.

One: permutations and combinations.

Solution: Select first and then arrange.

I think you should first understand the two big formulas that permutate and combine, multiplication and addition. Multiplication is step-by-step, and addition is classification. That's what I said about how you want to divide it, whether it's a step or a classification.

The key is to read the book, the textbook is basic, but I am the second phase of the course reform, some of the content jumped too fast, I also thought it was difficult at first, but it was good to do more, my trick is [choose first and then arrange].

In fact, permutations and combinations are also related to logical reasoning, I don't like the permutations and combinations that those people say that there are only 4 points in the college entrance examination room, but 4 points are also 4 points! Oh, permutations and combinations are related to probability, if permutations and combinations are not learned well, and there is a problem with probability, then it's not just 4 points.

Example: (in the textbook) Arrange six different elements of ABCDEF in a column, where A is not in the first place, B is not at the end, how many ways are there to arrange it?

If nothing is counted, the total arrangement of the 6 elements is a(6,6), but it is done with the unlimited, a is in the first place, then there are a(5,5) kinds, in the same way, b is at the end, there are a(5,5) kinds, the two are added, don't think it's weird, there is one more, a has ab once, b is also, so subtract a a in the first place, b at the end. It's a(4,4).

a(6,6)-(2*a(5,5)-a(4,4)=504 species.

Method 2: (It's too cumbersome, do it on the front, the top is the tail).

According to the elements, it is very cumbersome and easy to make mistakes, if you want to know, just ask, I don't come out here.

The new textbook weakens this chapter: Main methods: interpolation method, special element analysis method, partition method, division method, elimination method, etc.

There are a44 of them.

i.e. 4 3 2 1 = 24.

The sum of each four-digit digit is 10+x

There are 24, then 24 (10 + x) = 288

Get x=2

1) 4!*7-3!=162 (explanation: in addition to these 3 numbers, there are 7 numbers from 0 to 9, and finally subtract the three digits starting with 0).

2) 288 24 = 12 12-1-4-5 = 2 x is 2

2 elderly people are next to each other but not at both ends, the elderly have 4*2=8 ways, and the rest of the volunteers have A5 5=120 ways.

So there are a total of 8*120=960 kinds of arrangement.

With the interpolated method, the arrangement of 5 volunteers may have the method in A55; And there are only 2 ways for the elderly to be together.

And then it's like a hole in five circles. If you are not on either side, there are only four empty cards. Therefore, it can only be a55*a22*4=120*2*4=960

1 First divide two people who can speak English into A22 cases, and then divide computer programmers, three people are divided into two departments, two people in one department, one person in one department, so the group is C32 and then divided into two departments is A22, and the remaining three people are also two people in one department, and one person in one department is C32 A22, and the result is A22 C32 A22 C32 A22 = 72

2 first plant the middle piece, there are A41 choices, now it is to use the remaining four pieces of three colors, starting from the lower left corner, there are A31 choices, and then it is necessary to classify, the first category: the lower right corner and the upper left corner of the same color, the lower right corner has A21 choices, and the upper right corner also has A21 choices. Category II:

The lower right corner is not the same color as the upper left corner, there are a21 choices in the lower right corner, there is only one choice in the upper right corner (the same color as the lower left corner), and there are a21 choices in the upper left corner, so the result is a41 a31 a21 a21 + a41 a31 a21 a21 a21 = 96 kinds.

1 There are two English speakers and two more.

And then three computer programmers are assigned to C2, three of which are divided into two divisions, which are multiplied by A2, 2

Finally, the remaining three are allocated, i.e. a3,3

Multiply the three formulas.

2 Classification Divided into species 2 3 4 flowers to do is very simple.

I'm sorry, I will, but I can't type math symbols on the computer. I can only tell you the answer. Question 1; 72.

The second question is divided into two categories. The answer is 72

Take any 2,3,4,5 numbers according to size, and choose a method for each adjacent two numbers: c(5,2)*(2-1)+c(5,3)*(3-1)+c(5,4)*(4-1)+c(5,5)*(5-1).

When the largest number in a is 1, the choice type of A is 1, and the choice type of B is 15, a total of 1*15=15

So there are 49 in total.

b has only 1 element, and a has 4!/3!1!+4!/2!2!+4!/1!3!+4!/4!

1b2 elements 1 a has 3!/2!1!+3!/1!2!+3!/3!

1b3 elements 1 a has 2!/1!1!+2!/2!

1b4 elements 1 a has 1

There are a total of (4!.)/3!1!

I don't remember very well, it looks like I miscalculated just now.

When the minimum number of B is 5, A has 1 selection, B has 2 4-1 = 15 selections, when the minimum number of B is 4, A has 2 selections, B has 2 3-1=7 selections, when the minimum number of B is 3, A has 2 2=4 selections, B has 2 2-1=3 selections, when the minimum number B is 2, A has 2 3=8 selections, B has 1 selection, B has 1 selection, B has 2 4=16 selections, B has 0 selections, there are 1 15+2 7+4 3+8 1 49 choices.

First, when the largest number in a is 1, then b can be 2, 3, 4, 5, then it is the fourth power of 2 minus one is 15

Second, when the largest number in a is 2, then a has 2 in addition to the first case, b can be 3, 4, 5, then it is (2 to the third power minus one) multiplied by 2 is 14

Third, when the largest number in a is 3, then a has 4 kinds in addition to the first and second cases, b can be 4, 5 then (2 to the second power minus one) multiplied by 4 is 12

When the largest number in the fourth type of A is 4, then A has 8 kinds in addition to the above three cases, and B can only be 5, which is 8

There are 15 + 14 + 12 + 8 = 49 different selection methods.