Arrangements of children Can somebody please double check my work?
$n$ children must be arranged in a line.  $k$ pairs of children want to be next to each other, and each member of the pair will be unhappy if they are not next to each other.  The other children don't care where they are (i.e., they will be happy anywhere)
If all the arrangements are equally likely, what is the probability that all the children will be happy?
There are $(n - k)$ bins in which a pair or a happy kid can be placed (considering each as a distinguishable "atom").  So there are $(n - k)!$ ways to arrange the pairs and kids.  Each pair can be arranged in $2$ ways, so there are $2^k$ ways to arrange all the pairs (considering them as "compound" entities).  So there are $2^k (n - k)!$ arrangements which make the kids happy, so $p = \frac{2^k}{P(n,k)}$
I am asking for help because either my notes were wrong when I made them or I am wrong now.  I'm not confident that I can tell.  (In particular, they differ by a factor of $k!$, so the answer in my notes is basically $\frac{2^k}{\binom{n}{k}}$
 A: 
$n$  children must be arranged in a line. $k$ pairs of children want to be next to each other, and each member of the pair will be unhappy if they are not next to each other. The other children don't care where they are (i.e., they will be happy anywhere)
If all the arrangements are equally likely, what is the probability that all the children will be happy?

Assumption: None of the children want to be in more than one pair.
Count the ways to arrange $(n-k)$ distinct groups; consisting of $(n-2k)$ single children and $(k)$ pairs; then multiply by the ways to arrange children within each group (ie: $\times 2!$ for each pair, and $\times 1!$ for each singleton).  That is: $2^k(n-k)!$ happy arrangements.
Count the total ways to arrange $n$ children.  That is $n!$ arrangements.
The probability of a happy arrangement is: $\dfrac{2^k(n-k)!}{n!}$
A: We need to choose where the leftmost of each pair of fussy children will sit. Write down $n-k$ stars, where the other $n-k$ children will sit, like this
$$\ast\quad\ast\quad\ast\quad\ast\quad\ast\quad\ast\quad\ast\quad\ast\quad\ast\quad\ast$$
There are $n-k$ positions where these leftmost children can sit: In one of the $n-k-1$ gaps between stars, or at the left end. There are $\binom{n-k}{k}$ ways to choose these positions. 
These positions can be filled by children from the fussy pairs in $k!2^k$ ways. There is then only $1$ way to fill the positions to the right of the chosen positions. The rest of the positions can be filled with unfussy children in $(n-2k)!$ ways. So the required probability is 
$$\dfrac{\binom{n-k}{k}k!\,2^k(n-2k)!}{n!}.$$ 
The expression can be simplified in various ways. In particular, since $\binom{n-k}{k}=\frac{(n-k)!}{k!(n-2k)!}$, it simplifies to $\frac{(n-k)!2^k}{n!}$. 
