A permutation question, a permutation question

When a2 = 9, there are 8 * 9 = 72 species.

When a2=8, there are 7*8=56 species.

When a2 = 7, there are 6 * 7 = 42 species.

When a2=6, there are 5*6=30 species.

When a2=5, there are 4*5=20 species.

When a2=4, there are 3*4=12 species.

When a2=3, there are 2*3=6 types.

When a2=2, there are 1*2=2 types.

There are 240 species in total.

The key to solving this problem is not duplication or omission, there are two cases, a1=a3 and unequal a2=1 0 a2=2, 1 a2=3, 2*2-2=2 a2=4, there are 3*3-3=6 a2=5,4*4-4=12 a2=6,5*5-5=20 a2=7,6*6-6=30 a2=8,7*7-7=42 a2=9,8*8-8=56 1+2+6+12+20+30+42+56=169

1- None. 2- 1*2=2 types.

3- 2*3=6 types.

4 - 3 * 4 = 12 species.

5 - 4 * 5 = 20 species.

6- 5*6=30 species.

7 - 6 * 7 = 42 species.

8- 7*8=56 species.

9- 8*9=72 species.

There are 240 species in total.

Take two to stand out, c2 n way to take it, back and forth is not the same ticket, that is, a2 n = a2 2 * c2 n

Let each station be a point, that is, n points. For every two points are ticketed, that is, C2 N tickets, and Shanghai to Beijing and Beijing to Shanghai are not the same ticket, then it is necessary to multiply 2 to A2 2, so it is A2 2 * C2 N = A2 N kind of tickets.

There are n stations, take the first site, there are n ways to take it, and then select the second site, there are n-1 ways to take it; The two stations are selected to have a total of n(n-1), because they contain the successive stations, so it already contains the meaning of going back and forth, which is the final result, a2 n=n(n-1).

Discuss on a case-by-case basis.

If there are two different numbers for this four-digit number (e.g. 1212), there are c4, 2 possibilities.

If the four-digit number has three different numbers (e.g. 1213), there are 4*(2+2)=16 possibilities.

If the four-digit number has four different numbers (e.g. 1234), there are a3 and 3 = 6 possibilities.

So there are 28 numbers in total.

a=1, the remaining three numbers can be changed, respectively, a44*a33=144

If the remaining main force is allowed to be arranged in 2,4: first decide 1,3,5, and then arrange the 2,4 position with two of the remaining 5 people, then it is 4*3*2*(8-3)*(8-4)=480 Each factor represents the number of selectable species for each step, using the principle of step-by-step multiplication and counting;

If the remaining main force is not allowed to go up, then the 2nd and 4th positions are missing a choice, which is 4*3*2*4*3=288 or the principle of step-by-step multiplication counting.

With all due respect, the landlord didn't express my question clearly, I don't know if the 2,4 position will let the main force go up, but both answers are here!

According to the title, there must be 3 main players to participate, that is, there are p(4,3) possibilities, and there are 5 remaining players to select 2 players, because there are only two positions, so the full arrangement, that is, p(5,2), so there are a total of p(4,3)*p(5,2) methods.

3+2+2=7, box one has 7 balls, 2+3+4=9, box two has 9 balls, (1).

The probability of getting a blue ball from a box, c(3,1) c(7,1)=3 7, the probability of getting a blue ball from a box, c(2,1) c(9,1)=2 9, can not get the probability of a blue ball, (1-3 7) (1-2 9)=4 7 7 9=4 9, at least one probability of getting a blue ball, 1-4 9=5 9.

2）[c(3,1)/c(7,1)]×c(4,1)/c(9,1)]+c(2,1)/c(7,1)]×c(2,1)/c(9,1)]

16/63，3）16/63÷5/9=16/35，

The first floor is correct, but it is too complicated, everything else is wrong, the landlord should not be misled, first divide the 9 computers into three parts, to ensure that each of the three copies has at least two, there can be 2-2-5 or 2-3-4 or 3-3-3 three divisions, and then put the divided three parts in three positions (that is, three schools) Total: c31 + a33 + 1 = 3 + 6 + 1 = 10 kinds If this question does not require at least two for each school, the simplest is the partition method, that is, c83 = 56 kinds.

Hello! In fact, the meaning of this question is that three computers are randomly assigned to three elementary schools, and the number of allocations is uncertain.

It has become the first step: the combination problem, 3 computers can have three combinations of three numbers, note that the combination is not a permutation.

Then there is the problem of three permutations of three numbers, and there is 111 is the same number arrangement, in no particular order, there is only 1 kind, and 003 combination has two numbers of the same, so it is only 1 2, that is, 3 kinds, and 012 is three permutations, that is, 6 kinds, so the total number is:

1 + 3 + 6 = 10 species.

As for the algorithm of permutations and combinations, in fact, here is the simplest number of permutations and combinations, and oral arithmetic can even be said to be known, so there is no need to talk about it. At most, p3=3! =3*2*1=6。

First of all, it must meet the needs of at least two units per school.

Suppose the three schools are A, B, and C.

Then: A, B, C.

And so on: the number of types of delivery is: 14.

Three, two in each school, is six, the remaining three are divided into three schools, and the rest is counted on your fingers.

There should be only 6 ways to send it.

Xiao Ming: a(1,4);

Xiaoli: a(1,4);

Xiaoqiang: a(1,4).

The principle of multiplication, we get 4 4 4 = 64

3 4 There are 3 possibilities for each task, because it is not stated that everyone can complete at least one or something.

After first drawing a white ball in the field, there are 1 white ball and 2 black balls left, so the probability of touching another white ball is 1 3

Because the title has already said that first touch a white ball Jane stupid staring, this means that touching a block and a white ball has already happened, and there is no need to calculate it.

It's just that the topic is misunderstood.

Right. The first time it was determined was that the white ball was touched. So the next thing is to touch it between a white ball and two black balls, and the probability of the white ball is that it is not burned 1 3

This problem is a step-by-step counting problem, first arrange the tour method of city A, A and B can not participate in the tour method of city A, there are 4 ways to choose, and then look at the remaining three, you can choose any of the remaining five people, and get the result according to the principle of step-by-step counting

Answer: There are 4 ways to arrange a tour of city A first, and then there are 5 ways to visit city B, and then there are 4 ways to visit city C, and then there are 3 ways to visit city D According to the principle of step-by-step counting, there are 4 5 4 3 = 240 different options.