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.
 A: 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.
A: Let the children be $ABCDEFGH$ around the circle.  As the animals are different, I assume we consider rotations distinct.  First seat $A$ in one of the $8$ seats.  We have $6$ choices of children (all but $E$) that we can seat opposite $A$.  Now seat $E$ in one of the remaining seats, which is $6$ choices.  This is $8 \cdot 6 \cdot 6=288$ choices so far.  There are now two cases to consider.  
First, we can seat the partner of the child facing $A$ across from $E$, which is unique.  For example, if $C$ is facing $A$, we would seat $G$ facing $E$.  We are left with two pairs of children who were facing each other previously, so there are two ways to pair them up.  We have two choices for the pair of seats for the first pair, and two ways to orient each pair for a total of $16$ configurations.
Second, we can seat somebody else facing $E$, which gives $4$ choices.  This leaves two children who were facing each other previously and two whose mates are already taken.  We will next seat the earlier letter of the remaining pair in one of the $4$ seats.  We have two choices to face this child, and two arrangements of the remaining two children, for a total of $4 \cdot 4 \cdot 2 \cdot 2=64$ configurations.  
Putting it all together, we have $288(16+64)=23,040$ ways to scramble them.  If rotations are not distinct, we divide by $8$ to get $2,880$ ways to arrange them.
