How many ways can 8 children facing each other in a circle change seats so that each faces a different child. Need some help with this problem.  A carousel has eight seats each representing a different animal. Eight children are seated on the carousel but facing inward, so each child is staring at another. In how many ways can they change seats so that each faces a different child.
Was thinking P(8,8) for the total positions, but not sure where to go from there.

I calculate 23040 ways.
The way I see the problem, we have to assign 4 pairs of children (sitting opposite to each other) to 4 distinct slots. 
First let us calculate the number of way to seat the children, if we fix the pairs of oppositing children. Then the pairs could be assigned in $4! = 24$ ways to the slots. Since each of the pairs can be flipped, we have to multiply by $2^4 = 16$. So for fixed pairs of children, we have $4!*2^4 = 384$ ways to place them.
Now let us see, how many possibilities there are, to pair the children. I will call the children a,b,c,d,e,f,g,h from here on. Let us assume, that the start configuration is $(a,b),(c,d),(e,f),(g,h)$. So we are looking for the number all sets of such four pairs where none of the above is included. There are $6*5 = 30$ ways to build pairs of the form $(a,x),(b,y)$ with $x \in \{c,d,e,f,g,h\}, y \in \{c,d,e,f,g,h\} \setminus \{x\}$. Now there are four children left and at least two of them included in a forbidden pair, so they cannot be matched, which leaves only $2$ possibilities to match the other two pairs. So the number of desired sets is $6*5*2 = 60$.
In total we get $384 * 60 = 23040$ ways to seat the children in the desired way.
Question: is there a general combinatorical formula, which could be used to get to the 60? I do not see, how I could model this problem in a way, such that I could make use of one I know.
