$k$ kids, father and mother, it is forbidden to sit two chosen kids near to each other, How many ways there are to arrange them in line? I have the following question : We have $k$ kids, father and mother, it is forbidden to sit two chosen kids near to each other. 
How many ways there are to arrange them in line?
This is what I thought :
Total ways : $(k+2)!$
The total number of possibilities that the two kids sit next to each other : $2!*(k+1)!$ 
So this is what I got $$(k+2)!-2!*(k+1)!$$
But for some reason in the book they got :
$$((k+1)!*2!*k!)*(k+2)$$
Thank you.
 A: Your answer is correct.  The book's answer is not.  
Here is an alternate method of approaching the problem that the authors of the book may have had in mind:
Line up everybody except the two particular children who are not to sit next to each other.  Since there are $k + 2$ people and two are excluded, there are $k!$ ways to do this.  This creates $k + 1$ spaces in which the two particular children can be placed, $k - 1$ spaces between successive people and the two spaces at the ends of the row.  Since the two particular children are not to sit next to each other, choose two of these $k + 1$ spaces in which to place them.  They can be placed in these spaces in these spaces in $2!$ orders.  Hence, the number of possible seating arrangements is 
$$k! \cdot \binom{k + 1}{2} \cdot 2! = k! \cdot \frac{(k + 1)!}{2!(k - 1)!} \cdot 2! = k(k - 1)! \cdot \frac{(k + 1)!}{2!(k - 1)!} \cdot 2! = k(k + 1)!$$
Notice that this agrees with your answer since 
$$(k + 2)! - 2(k + 1)! = (k + 2)(k + 1)! - 2(k + 1)! = (k + 2 - 2)(k + 1)! = k(k + 1)!$$
A: Here's a better and seemingly easier method. First put the father, the mother and the rest of the childeren in the line. There are $k!$ ways to do that. Now between there are $k+1$ empty spaces (you need to count the ones on the edges too). So in order to sit exactly one child in those places reduces to choosing $2$ different places out of the $k+1$ places, which is $\binom{k+1}{2}$. But the two children can permutate so there are $\binom{k+1}{2}2!$ ways to sit them. at the end there are:
$$2!\cdot k!\binom{k+1}{2} \quad \text{ways to line them all}$$
