Some math problems during the winter vacation of the third year of high school, about permutations a

Dizzy, I made it to 3:30 last night and forgot to send it.

1、c5 1*c4 2*c2 2=30。If the division is 1, 2, 2, take one of the 5 first, then 2 from the remaining 4, and the last 2 will be the same.

2、c6 1*c5 2*c3 3=60。The principle is the same as above, first take 1 person into the second carriage (the first car is unmanned), the remaining 5 people take 2 people into the 3rd section, and the last 3 people do not need to be said.

2。A and B are not in a group and A and B are not in a group, respectively, there is 1 and a half cases, I understand.

4. No, to tell you the truth, I'm afraid of misleading you. My idea is to pay attention to the 4 corners, but I don't think the whole question is given, what if I throw it elsewhere? How do you calculate? I hope to ask the teacher for help and see what they say.

5、（c7 2*a2 2*a3 3*a5 5）/a10 10=120。Tie up the Martians and count them as a person, together with the earthlings to do a full arrangement, a total of 6, so there are 7 empty spaces that can be inserted (even the head and tail), of which 2 are selected to insert the Mercury people, they will never be lined up together, c72*a22, in addition, the Martians and the earthlings can be arranged internally so a33*a55, and then divide by the total score a1010.

3*c1 2*a2 2+10*a3 3）/c10 3=3/5。

Indirect method. Numbers divisible by 3, the sum of the digits is a multiple of 3, and the possible cases are listed separately: 0 and above:

012，015，018；Not 0 starts: 123,126,129,135,138,147,156,159, a total of 3 kinds at the beginning, discuss separately, first choose 1 from the 2 numbers that are not 0 as hundreds, and then the remaining 2 numbers a2 2 can be, there are 3 numbers on 3 * c2 1 * a2 2. There are 10 kinds of numbers that do not start with 0, you can use 10*a3 3, and then divide by the total score c10 3, and subtract it with 1.

7、（12/15）*（11/14）*（10/13）*（9/12）*（1/3）=11/94。Look at the formula, you know, the first 4 times are all white balls, each time the probability is decreasing because it is not put back, each time the numerator and denominator are reduced by 1, and the 5th time you can take 1 of the 3 red balls.

8、p（a）=。I don't have a calculator, so I won't help you calculate it Anyway, the odds of Superman winning are greater, do you see it? This question is about you hitting me once and I hit again, you just need to pay attention to Superman and definitely not be in the first 4 times. That's it.

9. What does p(non-b a)=1 mean? I haven't reviewed it either, hehe, listen to the teacher

By the way, I also happened to be in the second semester of my third year of high school, so I worked hard together, and I had the opportunity to talk about mathematics or other subjects together, thank you.

I'm passing by, but remind you, I hope you take a serious look, the types of these questions I have seen in the college entrance examination, not difficult problems, medium questions, some even simple questions, for your questions: afraid that the answer is wrong, take the first question as an example, c52*c32*c11 This formula is still listed, but it is estimated that many students are puzzled: do you need to multiply it by an a33?

So when doing this kind of question, you have to think about what the event is, and make a list on the paper: 5 people take two, 3 people go to two, and the last person is the last one; At this point, this is done, so there is no need to arrange it again, because there is no specific class between the classes themselves.

Therefore, the problem of permutations and combinations is: first, to see what kind of event it is; The second is: this matter is not over, if it is over, don't add to it.

Since it's a holiday homework, you should think about these questions yourself, even if there is no answer, the wrong is wrong, and you are not confident in your heart, but it is affected in the examination room, and you always think about whether it is right or not? Let someone else do it, but that's someone else's ability, and it doesn't fall to you in the end. This part of the mathematics should be --- and rigorous; You also need to have --- experience, so do more practice.

I'm not here to ask you questions.,But I suggest you these.,I hope that after waiting for your questions.,Don't do it in the future.,It's not helpful for your learning.,Answers,Call **Ask the teacher.,Wait for the teacher to announce the answer at the beginning of school.,Isn't it all the same.。 The study method is very important, you have a problem here, you have to think about when you are revising, whether certain measures make sense?

Question 1: First of all, there is one person in each class, and there are C5-3*A3-3 kinds; Then there are two people in a class, and the remaining one person has c5-2*c3-1*c2-1 species; Finally, there are two classes with two people, C5-2*C3-2 species; It adds up to 150 kinds.

There is no problem with the second question.

Question 5: It may be that your answer is wrong, I also calculated 1 20 Question 7: You may wish to set the fifth draw is the red ball, then the first four draws are:

Four white, three white and one red, two white and two red. So it's 12 15*11 14*10 13*9 12+4*(3*12*11*10) (15*14*13*12)+6*(3*2*12*11) (15*14*13*12).

Question 1 c5 2*c3 2=30 30*3=9oI hope to understand.

I didn't see the problem in the second question.

Question 3: 50% 50% chance of A to group A, 50% to group B, the same for B. Therefore, the probability that A and B are in the same group is 50%*50%*2=50%.

I don't have time, I'll go back to something else when I have time.

Since it is a holiday homework, you should think about these questions yourself, even if there are no answers, you are wrong.

If you haven't reviewed it, won't you, I'm depressed, are you studying liberal arts or science!

I've done this question before, so I'll make it easy to understand.

Let's say four teachers are each a b c d and the class they teach is a b c d

First of all, Teacher A chooses, and he has three options: B, C, and D.

If Teacher A chooses C, then let Teacher C choose next, and Teacher C also has three choices (A, B, D).

If Teacher C chooses A, then B and D have to choose D separately B If Teacher C chooses B, then B and D have to choose D A separately If Teacher C chooses D, then B and D have to choose A B means that the first two teachers have three situations, and the remaining two teachers must have only one situation (you can draw a picture on paper, it is easy to understand).

Hence 3 3 = 9

Suppose four teachers A, B, C, and D teach in four classes, then during the exam, each teacher invigilates one class.

4 3 2 1 = 24 ways, and the situation of the teacher in his own class in these 24 combinations is:

There is only one teacher in his class: assuming that A is in class 1, then class 2 has class C, and class D invigilation corresponds to one type of invigilation.

Then the total is 4 2 = 8.

There are two teachers in their own class: if A and B are in their own class, then C and D are not in their own class and there is only one situation.

Then consider which two teachers are in their own class: 4 3 2 or 3+2+1=6 (permutations) There are three teachers in their own class: the fourth teacher must be in their own class, so it is 1.

In the end, there is a total method of proctoring.

24-8-6-1 = 9 (species).

There are three teachers invigilating the class, which is 4 for the whole class, and there is only 1 type of invigilation in this class, so there are 24-8-6-1 = 9 kinds of invigilation methods.

The teacher of a class can have three ways to choose, and the teacher of the class he chooses also has three ways to choose, and the remaining two teachers each have only one way to choose, so there are a total of 3x3x1x1=9 ways to choose.

You're all wrong, the first case doesn't say what kind of sugar it is.

You can give chocolate to the first person and marshmallows to the second person ......So there should be c(3,xp(5,+c(2xp(5=240 kinds, no problem, the second question is 2156

1. Consider driving by one person first, then there are a(1,3)=3 kinds of arrangement, so that there are 4 people and three jobs left, and the rest can be arranged at will, considering that every job has to be done, then there must be a job that needs 2 people, you can choose 2 people first, there are c(2,4)=6 kinds, so that 4 people can be regarded as 3 people, and the arrangement is a(3,3)=6, then the total arrangement scheme is a(1,3) c(2,4) a(3,3) = 72 kinds.

2. If there are 2 people driving, there are c(2,3)=3 kinds, and the remaining 3 people do 3 jobs, it is a(3,3) kinds, then the total arrangement scheme is c(2,3) a(3,3)=18 kinds.

72 18 = 90 species.

That's what I thought.,I don't know if it's right.。。 In this way, the first step is to determine who will be the driver, because A and B can't be the driver, so there are three people left, that is, choose one from the three, because there is no need to care about the arrangement, so it is good to use the combination C (in fact, I'm not sure whether to use P or C here).

The second part is to randomly select three of the remaining four people to do the remaining three jobs, which should actually be calculated in the P arrangement.

I did the math and it seems that the answer is not quite right. Just give LZ an idea, I hope it can help

First A and B, there are two situations, the same work and the different;

For the same job, there are 3 options, and then there are 3 people left with 3 jobs, all arranged, a33=6;18 species in total;

For different jobs, there are 6 options, and the remaining a33, = 6, 36;

You get 54

How could it be 126

5 people, engaged in one of the four jobs, then it is decided that there are 2 of the same, and the others are one each;

Using the tree diagram, even if A and B can drive, there are only 120 possibilities, and the answer is wrong. I'm counting 72.

1. Determine the color of point E first, there are c(1,4)=4 kinds, then point a, d can only be selected from the remaining 3 colors, there are a(2,3)=6 kinds;

2. In the case that E has already determined the color, the color of point F can only be selected from the remaining 3 colors.

If F and D are different colors, that is, the color of F is A(1,3)=3, then B's choice is A(1,2)=2, please note that C is different from B, F, and D.

At this time, the filling schemes are: c(1,4) a(2,3) a(1,3) a(1,2)=144 kinds;

If f and d are the same color, then f does not need to be selected, and the method of b has a(1,2), considering that c should be different from the colors of b, f, and d (at this time, these three points share two colors), then c has c(1,2).

At this time, the filling schemes are: c(1,4) a(2,3) c(1,2) a(1,2) = 96 types.

The total number of coating schemes is: 144 96 = 240 types.

Since 8 is the largest number, the ordinal number of 8 is 2, which means that 8 is in the 3rd position, as shown below.

7 is second only to 8, and the ordinal number of 7 is 3, so 7 can only be ranked in the 5th position, as shown below.

The ordinal number of 5 is 3, but there is a position greater than 5 in 6 that is not determined.

If 6 is to the right of 5, then the 3 digits in the first, second, and fourth digits must be smaller than 5, so 5 is in 6th place.

In this case, 6 can be ranked in the seventh or eighth position, and the remaining numbers can be inserted into the remaining air in full array, so the number of species is 2*4!=48

If 6 is to the right of 5, then 5 is 7th.

In this case, 6 can be ranked in the first, second, fourth, and sixth positions, and the remaining numbers can be fully arranged and inserted into the rest of the air, so the number of species is 4*4!=96

So the total is 48 + 96 = 144

There is also a way, after determining the position of 7,8.

Since there is no limit to 6, there are 6 possibilities to fill in casually.

When 6 is filled, the position of 5 is determined uniquely, and the remaining 4 numbers are all arranged.

So it's 6*4!=144

Further, at the beginning, let the unrestricted 1, 2, 3, and 4 all be arranged, insert 5 without selection due to position constraints, then insert 6, then 7 restricted, and then 8 restricted.

So compared to the full arrangement of 8 numbers, there is less number of permutations of restricted numbers.

So it's 8!/(8*7*5)=144

With this formula, even if the question is changed slightly, the answer can be quickly obtained.

Analysis: ordinal number, i.e. the number of numbers on the left that are smaller than it; The ordinal number of 8 is 2, that is, there are only two numbers smaller than it in front of 8 (left), because the other numbers in 1 8 are smaller than it, so there can only be two on the left side of 8, that is, the row position is in the third place. If the ordinal number of 7 is 3, then consider that the first two are smaller than 7, and the third digit 8 is not counted, and one more number is needed, then 7 ranks fifth.

The ordinal number of 5 is 3, and if it is the same as 7, it may be directly after 7, i.e., the sixth place; It may also exist between 7 and 5, not counting 6 in the ordinal number (because it is greater than 5), at which point 6 is sixth and 5 is seventh.

The calculation is as follows: 8 row third, 7 fifth, 5 sixth; At this time, make a full arrangement of the remaining 5 numbers: 120 or: 8 is third, 7 is fifth, 6 is sixth, 5 is seventh; At this point, all the remaining 4 numbers are arranged: 24

Add it up to: 144