Eight people lined up. If there is just 1 person between A and B, and C and D are not adjacent, how many arrangements are there?

This problem can be solved by classified discussion and bundling.

Let's line up A and B first, then insert one of the remaining people between A and B, and then tie one person between A and B together and fill it with others (this will ensure that there is only one person between A and B).

(1) If the person between A and B is not C and D,

First of all, there are two kinds of A (2 2,2) = C (4, 1)=4 in line A and line B, and one of the remaining 4 people is inserted between A and B.. Then the whole and the other three people except C and D are completely arranged with a (4,4) = 24 kinds, and then C and D are inserted into four elements including A in five spaces on both sides by interpolation.

Then, there are always 2 * * * 4 * 24 * 20 = 3840 such cases.

(2) If the person between A and B is C or D.

There are two kinds of A (2,2) in row A and row B, two kinds of C(2,1) for one person in row C and row D, and 720 kinds of A (6,6) for all and the other five people.

Then, there are always 2 * * * 2 * 720 = 2880 kinds of this situation.

To sum up, there are 3840+2880=6720 schools in total.

I hope you can understand, because it is not easy to subscript and so on. , so it is represented by (,). The comma is preceded by a subscript, and the comma is a superscript.

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