A permutation problem, a permutation and combination problem

The denominator represents the number of permutations of 20 heroes out of 90 heroes.

The numerator means that the heroes you pick are the number of permutations you don't use.

The whole score represents the probability that you didn't hit a hero you can use, and subtracting it from 1 gives you the probability of hitting a hero you can use.

This kind of problem is relatively simple, but you can see from everyday life that the spirit of mathematics is commendable, and you, big, good. Hehe.

The heroes you will use make out of the total heroes is 4 90

Pick 20 heroes.

So the probability is 4 90 * 20 = 8 9

I've lost it for too long to remember.

But if you don't remember, the odds of being upstairs that big are definitely not correct.

Make a soy sauce.

Choose 4 out of 20 c(20)4

Choose 4 out of 90 c(90)4

Probability = c(20)4 c(90)4=20*19*18*17 90*89*88*87

By the way, LZ is so strong that he can actually think about these problems from Dota.

Admire, admire.

Pick 20 out of 90. There are a total of C90 20 = 50980740277700939310 selection methods. If you get all of the 20 and you can't, you will choose 86 out of 90.

is c86 20 = 18293741700978245355Divide c86 20 by c90 20 to get 20 and not have a single probability that you will use it. Subtract the number by 1 to get the probability...

Get the probability that you will use the draw. Complete.

This kind of question should not be done in the same way as yours, because you will double-count certain DAO cases.

In addition, it is definitely not right to go back to c12 1* c8 1* c18 3, and you should also answer c12 1* c8 1* c15 3, of course, it is also wrong.

Example: Inner ACD and Outer E

C12 1, choose A, C8 1, choose E, C15 3, choose BCDC12 1, choose B, C8 1, choose E, C15 3, select ACD has been repeated

So it should be discussed in categories:

Inner 1, outer 4: c12 (1) * c8 (4) = 12 * 70 = 840 inner 2, outer 3: c12 (2) * c8 (3) = 66 * 56 = 3696 inner 3, outer 2:

C12 (3) * C8 (2) = 220 * 28 = 6160 inner 4, outer 1: C12 (4) * C8 (1) = 495 * 8 = 3960 add: 840 + 3696 + 6160 + 3960 = 14656 If you don't understand, please ask.

With reverse thinking, all internal medicine cases = c12 * 5, all surgical cases = c8 * 5, all cases = c20 * 5, so the result is c20 * 5 - c12 * 5 - c8 * 5.

The conclusion of this question is 6!s(10,6)=16435440.

where s(n,k) represents the second type of stirling number, and its combined meaning is: divide the n-yuan set into k non-empty subsets, and the order between the subsets is not counted, and the resulting division number is s(

In this problem, 10 people are equivalent to a 10 yuan set, and 6 stations are equivalent to 6 non-empty subsets. Note that there is a difference between the stations, so the conclusion of this question is 6!s(10,6).

Generally speaking, s ( does not have a closed form of expression, which means that the problem cannot be expressed in a very simple form.

The recursive formula s(n,k)=s(n-1,k-1)+ks(n-1,k) and the initial value s(n,1)=s(k,k)=1 are commonly used in computers to find s(

This recursive proof is not difficult, and it is more interesting, let's talk about it below.

If a is taken from an n-element set, and if a monopolizes a set, then the problem becomes that the remaining n-1 numbers are divided into k-1 non-empty sets, and there is a s(n-1, k-1) division.

If there are other elements in the set where a is located, ignore a, and the remaining n-1 numbers are divided into a non-empty set, and there is s(n-1,k) division; When A is added, k different positions can be selected, so there is a ks (n-1, k) division at this time.

In summary, s(n,k)=s(n-1,k-1)+ks(n-1,k)

Another way to find s(n,k) is to use the repulsive principle, which is acceptable for the amount of calculation in this problem. Let's take this topic as an example.

If you don't take into account the condition that people get off at each stop, everyone has 6 options, and the conclusion is 6 10

That's obviously a lot, and at least one stop should be removed. First select one of the 6 stations and no one will play, and then let 10 people choose from the remaining five stations, a total of c(6,1)*(5 10) situations. The preliminary conclusion is 6 10-c (6, 1) * (5 10).

After careful analysis, the above process has been cut out a little. For example, when no one went down at the 1st and 2nd stations, no one at the first station was planed once when the top was removed, and then no one was planed at the second station. C(6,2)*(4 10) should be added

And so on, from the principle of repulsion, the conclusion should be:

6^10-c(6,1)*(5^10)+c(6,2)*(4^10)-c(6,3)*(3^10)+c(6,4)*(2^10)-c(6,5)*(1^10) (

To sum up, it is better to use the principle of repulsion in this problem, which can take into account the simplicity of calculation and the universality of ideas.

By the way, the method of "pengp0918" netizens is indeed feasible, and the calculated number is also correct (but an extra 1 is added in the last step). But that method is not ideologically universal. If K is large, there are too many situations to discuss, and it is too complicated.

However, the repulsive principle is different, as long as 10 and 6 are replaced by the general n and k, the above equation (*) can still find the answer.

1. 5 people at one station, 1 person at each other station: C10 (5) * 5 * 4 * 3 * 2 * 1 * 6 = 10 * 9 * 8 * 7 * 6 * 6 = 181440

2. 4 people at one station, 2 people at one station, and 1 person at each other station

c10（4）*c6（2）*4*3*2*1*6*5=10*9*8*7*6*5*6*5/2=2268000

3. 3 people at each of the two stations, and 1 person at each other station

c10（3）*c7（3）*4*3*2*1*c6（2）=10*10*9*8*7*6*5=1512000

4. 3 people at one station, 2 people at each station, and 1 person at each other station

c10（3）*c7（2）*c5（2）*3*2*1*c6（1）c5（2）=10*10*10*9*8*7*6*3=9072000

5. 2 people at each of the four stations, and 1 person at each other station

c10（2）*c8（2）*c6（2）*c4（2）*2*1*c6（2）=3402000

Therefore: 181440 + 2268000 + 1512000 + 9072000 + 3402000 = 16435441

There are cases where 16435441 get off the bus.

Step-by-step analysis:

First of all, at least one person at each station gets off, then the possibility of the first station is 10, the possibility of the second station is 9, and so on, the possibility of the sixth station is 5, in total: 10*9*8*7*6*5 possibilities.

In this way, the condition of "at least one person getting off at each station" is met Then, the remaining 4 people, for each person, there are 6 possibilities for him to get off, that is, at any of the 6 stations, so for the remaining 4 people, there are a total of: 6 4 possibilities.

So: getting off may be: 10*9*8*7*6*5*6 4=195955200

The first group puts one person at each station A total of 6 people are selected There are A (kinds, the second group is 4 people left, 6 stations are arbitrary, there are 6 4 kinds, and if any two people are set up, they are A and B.

If A is selected into one group, B is selected into two groups, and B is selected into one group, and A is selected into two groups, and both of them get off at the same time at any station, they will be counted twice.

Then there is a total of [a( species.

First of all, 6 stations with one person per station have A6 10 kinds;

There are 4 more people who can get off at any of the 6 stations;

No one has 6 options Total 6*6*6*6

A6 10*6*6*6*6=272,160 species.

1 1 1 1 1 1 1 1 1 1

Assuming that this is 10 people, we can first use the "plug-in method" to calculate that 10 people walk 6 stations to get off, and the scheme of "number of people" looks like this1|1|1|1|11|1111

It means that there are 1, 1, 1, 1, 2, 4 people get off from door 1 6 is equal to inserting 5 boards from 9 empty boards between the plates to indicate the person who gets off in one door c (9, 5) This means that we people get off the car in an orderly manner, so we can arrange the car door again, so that the position of the person getting off the car is arbitrary, and the order of getting off the bus in the same door will not affect the answer and repeat.

So the answer is.

c（9，5）*a(6,6)=90720

Upstairs upstairs is not evenly grouped for repetition is not taken into account.

Because there are six stops, and someone must get off at each stop, the number of people getting off can be divided into the following categories: 1, 1, 1, 1, 5

For the above combination operation, different passenger disembarkation combinations are obtained, and then the six stations are fully arranged to obtain the answer.

First of all, one person is arranged to get off the station at each station, and 10 people at 6 stations are A6 10, and there are 4 people, and there is a possibility of going down at each station, which is 6 4, and the result is A6 10 * 6 4

First, 6 out of 10 people are assigned to each station with a ( case, and then the remaining 4 people can have the following divisions:

Four people are under the same station, and one of the six stations is selected as C (

Four people are divided into equal groups, with two people in each group, and two stations are selected to assign two groups. c(2,4)c(2,2) divided by a(2,2) multiplies a(2,6) = 90

Four people are divided into two groups, a group of 3 people and a group of one person. c(3,4) times a(2,6) = 120151200 (6 + 90 + 120) = 32659200

It can be used in the plug-in board method:

There are a total of 10 people, and the insertion of five boards is divided into six parts, that is, six teams, to ensure that there are people at each station. 10 people, there are a total of 9 empty in the middle, so there are a total of C95 ways to arrange the board, and then arrange: A55 ways to arrange So, there are altogether.

c95 a55 = 10 * 9 * 8 * 7 * 6 = 15120 cases.

To give you a reasonable approach, I ran out of the program with a program!

The number of people getting off at each station may be 0 10 people in a total of 10 types, so the situation is 10 * 6 = 60 types.

First of all, let's take the ith child to get ai sugar, then a1 + a2 + a3 + a4 = 10, ai > = 0, that is, the number of solutions to the indefinite equation. Then we let bi=ai+1, then the equation is b1+b2+b3+b4=14, bi>=1.

We arrange the 14 pieces of sugar in turn, and then use 3 chopsticks to separate the 14 pieces of sugar, so that the sugar becomes four parts, and the first part is bi. Then there are as many ways to insert 14 pieces of sugar with chopsticks. Since there is at least one of each serving, chopsticks cannot be placed in the same empty space.

So there are a total of 14 pieces of sugar, there are 13 neutral positions, and 3 neutral blocks are taken out of them and inserted into the chopsticks. A total of c(3,13) = 286 divisions.

There are no 14 pieces of sugar in the picture, just simulate it, it's more beautiful)

This problem is called the "misalignment problem".

Derivation of the recursive formula for the misalignment problem:

When n numbered elements are placed in n numbered positions, the number of methods that do not correspond to the element number and the position number is represented by m(n), then m(n-1) means that n-1 numbered elements are placed in n-1 numbered positions, and the number of methods that do not correspond to each other, and so on.

The first step is to put the nth element in a position, such as position k, there are n-1 methods;

The second step is to put the element numbered k, and there are two cases to put it in position n, then, for the remaining n-1 element, since the k-th element is placed in position n, there is a m(n-2) method for the remaining n-2 elements; The k-th element does not put it in position n, and then there is a m(n-1) method for the n-1 element;

To sum up. m(n)=(n-1）[m(n-2）+m(n-1）]

In particular, m = 0, m = 1

The following is a general term formula derived from this recursive relation:

For convenience, let m(k)=k!n(k),(k=1,2，…，n)

Then n = 0, n = 1 2

n>=3, n!n(n)=(n-1）（n-1）！n(n-1）+(n-1）！n(n-2）

i.e. nn(n)=(n-1)n(n-1)+n(n-2).

So there is n(n)-n(n-1)=-[n(n-1)-n(n-2)] n=(-1 n)[-1 (n-1)][1 (n-2)]....1/3）[n⑵-n⑴]=(-1）^n/n!

So n(n-1)-n(n-2)=(-1) (n-1) (n-1)!

n⑵-n⑴=(-1）^2/2!

Add it up, you can get it.

n(n)=(-1）^2/2!+…1）^(n-1）/(n-1）！+1）^n/n!

Thus m(n)=n![(1）^2/2!+…1）^(n-1）/(n-1）！+1）^n/n!]

can be obtained. The staggered formula is m(n)=n! （1/2!-1/3!+…1）^n/n!)