Placing 5 people on chairs so that 2 people never sit together We have a family composed by 2 parents and 3 children that have to seat on 5 chairs (in line), two of the children don't want to stay together, how many are the ways that they can sit?
I was thinking about letting one of the 2 children sit first, now the second one can sit only on 3 chairs, the remaining parents and child now can sit on the other 3 chairs
So I came to that expression: 3*3! = 18 but I saw that this wasn't the correct answer...
If the chairs are in a line, then the calculation must take into account  where the first child sits.  There are two ways for him to take a seat on the end, then three ways for the other child to take a seat.  There are three ways for the first child to take a seat in the middle of the row, and then there are only two ways for the second child to take a seat.  So there are in all $$2\cdot3+3\cdot2=12$$ ways for the two of them to be seated.  There are then $3!$ ways to seat the rest of the family, so $72$ ways in all.
I think your idea is correct, you just miss some calculation.
I will assume here that the 5 chair are put in circle
So yes, let first choose the chair of one of the difficult child (5 choices, 4 chairs left)
Then the second difficult choose from the 2 chairs which are not together (1 chair as 2 neighbour, and one chair is already taken) (so 2 choices, and 3 chair left)
At least let the 3 other sit (3! choices)
So tu sums up, we have $5\times2\times 3! =60$ possibilities
Is this your expected answer, or did i miss something in the initial statement ?