6 people are lined up in a row, and 2 of the three people A, B, and C are lined up together, and the

First of all, we take 2 of the three ABCs and have C32=3*2 2=3, and then there is 1 left.

First of all, we just took the rest from the 3 and forgot to put the other 3 inside, obviously there are 4 positions that can be placed, and the other three are arranged.

p33 = 6 species, apparently now there are p41 * p33 = 24 species.

Now let's put the two that we just took in the middle, and obviously there are 5 places to put it.

Therefore, there are p33 * p41 * p51 * c32 = 6 * 4 * 5 * 3 = 360 kinds.

The way to completely separate the three ABCs is to arrange the three people with one in between:

Take one of the three ABC people as the head, and the permutations and combinations are 3*3*2*2*1*1=36 kinds.

Take one of the other three as the head, and the permutation and combination is 3*3*2*2*1*1=36 kinds.

There are 6*5*4*3*2*1=720 kinds of random arrangement of six people.

Then there are 720-36-36 = 648 ways to line up two people in ABC.

ABC is arranged in order of progress, and there are 6 cases in total.

In this way, ABC3 people are seen as the same people.

Then look at whether these 3 people are in a 2,1 row or 1,2 row, in both cases, and look at them as * and * The remaining 3 people are arranged together to have 6 people.

and * There are 3 cases when there is 1 person in the middle.

and * There are 2 cases when there are 2 people in the middle.

and * 1 case when there are 3 people in the middle.

There are a total of 6*2*6*(3+2+1)=432 cases.

1.Consider ab together.

2.Consider AC together.

3.Consider BC together.

The three possibilities add up.

Answer: 720

There are 6 in total! methods, i.e. 720 methods. The probability is 1 5 * 3 4 * 4 = 12 20, so it is 720 * 12 20 = 432 species.

ABCDE five people lined up in a row, for a total of 5 full rows.

Seed arrangement, i.e. 5!. And c in the middle, the other permutations except the middle should have 4! Seed arrangement. So it should be:

Seed arrangement.

Remove the arrangement of 4 people left = 4 3 2 = 24 (species) after removing c

Plus c total = 4 24 = 96 (species).

There are 5 students in a row, of which A and B are not lined together, and there are 96 different ways to arrange them.

The calculation process is as follows:

It can be calculated according to the title:

5 students in a row: a(5,5)=5 4 3 2 1=120 (species)AB is regarded as a person when they are lined up: a(4,4)=4 3 2 1=24 (species).

AB is not grouped together: 120-24 = 96 (species).

So A and B are not lined up together, and there are a total of 96 different arrangements.

5 students are arranged in a row, and there are a total of a(5,5)=5 4 3 2 1=120 kinds.

AB is arranged together, and AB can be regarded as a person, and the arrangement is: a(4,4)=4 3 2 1=24 kinds.

AB is not arranged together: 120-24 = 96 types.

It is possible to use the interpolation method. The first step is to arrange 3 people without a limit, and since it is a human arrangement, there is an order, so it is P33. Then in the second step, these three people include a total of 4 empty spaces at both ends, and you can arrange the two AB people into any two of the 4 airs, there is an order, so it is p42.

Multiply the two steps, so it is 72 kinds.

A in the first, B in 345; A in the second, B in 45; A in the third, B in the 15; A in the fourth, B in the 12; A is in the fifth, B is in the 123. So there are 12 types in total.

Divide A and B into the front row and the back row each of the two people are riser, 2!

Draw the remaining 4 people out of 2 people to the back row, and the remaining 2 people in the front row to raise the old, C4 (2).

Arrange the 3 people in the back row and the 3 people in the front row in full line, (3!)^2.

So the collapse of the hood, 2!*c4(2)*(3!2 = 2 * 6 * 36 = 432 species.

Answer]: B does not consider any restriction and there are 5 4 3 2 1 120 (species), if A and B are arranged together there are (4 3 2 1) 2 48 (species), and a total of 120 48 72 (Species of Destruction) is not arranged together.

Put A and B aside. The remaining 4 people lined up, 4!= 24 scenarios.

Counting the two ends of the 4-person queue, there are a total of 3 insertion schemes between two people, so that there are exactly two people in the middle of A and B. There are two types of order of A and B. Merge, there are 3 * 2 = 6 cases.

The above superposition calculation is a total of 24*6 = 144 different arrangements.

First, 9 people will be arranged, and there will be a total of 99 types of A. Then A, B, and C are all arranged, and there is A33In A99, every A33 disadvantage corresponds to a method that conforms to the rent and the old parts, so there is a total of A99 divided by A33=60480

The title is not quite accurate.

A few words should also be added: "The five persons selected must include A and B".

If that's the case, you can think about it, then choose three out of the five people other than A and B (the remaining two).

So there are only 10 options.

Then let 5 people, including A and B, line up.

There is a standing method of the queue.

5*4*3*2*1*10=1200 (species).