How many ways can the children be arrangged?

Question

There are $10$ children among whom there is one pair of a brother and a sister. They are not allowed to sit next to each other. How many ways can $10$ children be arranged in a straight line if the brother and sister can not sit next to each other?

I solved the answer which is $10!- (9!\times2!)$. This is right, however, another way of getting the same answer is $8\times9!$. I did not understand the second way of finding this answer. Can someone please explain why the answer is also $8\times9!$ ? Thanks for the help:)

Answer 1

As Alain said, you first place everyone except one of the brothers.

This is easier to visualize with an example:

Suppose the brothers are A and B and the remaining children are C,D,E,F,G,H,I,J.

You pick C,D,E,F,G,H,I,J and a brother B and rearrenge them as if you had only 9 seats:

(you have $9!$ ways to do this)

For example,

$$\text{F,C,I,D,G,H,J,E,B}$$

Now you need to see where can you insert A. You can insert it before F, between F and C, ... The only places where you cannot insert it are between E and B or after B (the same argument goes for any configuration of 9 children)

(you can insert A in $8$ places)

Thus you get $$9!\times8$$

Answer 2

Has said in the comments, a small manipulation could be used $$10!-2\times9!=(10-2)\times9!=8\times9!$$

The same could be achieved by a combinatiric argument. Since the brother and sister don't want to be together, start by placing everyone except one of them. If we remove the sister from the group, there is $9!$ to place the remaining children. There are $10$ places for the sister, two of which are next to her brother. So, it leave her with $8$ choices. $$8\times9!$$