Point out my Fallacy, in combinatorics problem, please. 
Five children sitting one behind the other in a five seater merry go round, decide to switch seats so that each child has a new companion in front. In how many ways can this be done?

My tries:
I try using IEP but didn't work, please point out a fallacy.
There are $4!$ without any restriction.
Let $p_1$ be the property such that one of them has the same companion in front, similarly for other properties, $p_2,p_3,p_4,p_5$, as well.
No. of way in which $p_1$ occur,
I used tie method, tie 1st one and the 2nd one, we remain with $3$, which along with tied one can be arranged in $3!$ (circular permutations). Similarly for other properties as well.
No of ways in which $p_1\cup p_2$ occur:
Now tie three consecutive people, so we remain with $2$, which along with tied peoples can be permuted in $2!$, similarly for other as well. Total results in $5\cdot 2!$. (need to tie consecutive from $5$, not any hence factor with $2$ is not ${{5}\choose{2}}$)
No. of ways in which $p_1\cup p_2\cup p_3$ occur:
Now, four consecutive people, we remain with $1$, which along with tied can be permuted in $1!$. Total results in $5\cdot 1!=5$
No. of ways in which $p_1\cup p_2\cup p_3\cup p_4$ occur:
Now we'll tie $5$ consecutive, so one way.
No. of ways in which $p_1\cup p_2\cup p_3\cup p_4\cup p_5$ occur:
will be same as No. of ways in which $p_1\cup_2\cup p_3\cup p_4$ occur $=1$
Exploiting IEP:
$$4!-(5\cdot 3!)+(5\cdot 2!)-(5\cdot 1!)+1-1=-1$$  
Where I over substracted !!!
Please Help.
 A: I do not really understand what you are counting exactly.
However, If $f(n)$ is the number of such rearrangements for $n$ children (counting those that can be transformed into another by rotating the merry go round), then we can find a recursion:
Let $n>3$ and the kids wear t-shirts numbered $1,\ldots,n$. In the original arrangement, the kids sit as $1\to 2\to\ldots\to n\to 1$.
In the $f(n)$ other arrangements, there is no case of $k\to k+1$, nor os there $n\to 1$.
Now suppose $n$ puts a head on his successor and leaves, i.e., we convert $\ldots\to a\to n\to b\to \ldots$ into $\ldots\to a\to \hat b\to \ldots$. If $b\ne a+1$, this is one of $f(n-1)$ valid situations with $n-1$ kids (note that it is not possible to have $a=n-1$ and $b=1$); additionally, some kid that is neither $1$ nor the successor of $n-1$ has a hat.
But if $b=a+1$, $a$ leaves with $n$, takes of his shirt and kids $a+1,\ldots, n-1$ switch shirts so that each decreases by one, and shirt $n-1$ remains unused. In the end, we have $\ldots\to \hat a\to\ldots$ where $n$ and the now shirtless kid left. So this is one of $f(n-2)$ valid arrangements and a kid $\ne 1$ and $\ne n-2$ (note that these two exceptions are different!) has a hat.
Because of the hat, we can undo these steps: If there are $n-1$ kids on the merry go round, $n$ simply inserts himself behind the hat-kid.
And if there are $n-2$ kids, then $n$ inserts behind the hat-kid with the shirtless kids behind him, the kids $\hat a,a+1,\ldots, n-2$ increase theri numbers while shirtless gets the then free $a$.
We conclude that
$$\tag1 f(n) = (n-3)f(n-1)+(n-4)f(n-2)\qquad\text{if }n>3.$$
Clearly, $f(1)=f(2)=0$, $f(3)=1$.
It is easy enough to use $(1)$ to find $f(5)$.
According to http://oeis.org/A000255, we have the surprising formula
$$f(n+3)=\left\lfloor\frac{n!(n+2)}{e}+\frac12 \right\rfloor.$$
A: If you have one reordering so that each child has a new companion in front, then there are 4 similar solutions. Theses can be obtained by rotating the childs any number of times by $2\pi/5$ about the center of the merry-go-round. Hence the answer is a multiple of 5.
Now I think it's simplest to just list the solutions. Start by choosing a rotation by assigning child 1 to, say, the 3rd seat. Then put in front of him 3, 4 or 5. Then behind him any but 5, whoever is left.
Put the last two childs however it can validly be done (0, 1 or 2 ways (in fact at least 1)) 
$\{4,2,1,3,5\}\ ,\ \{5,2,1,4,3\}\ ,\ \{3,2,1,5,4\}$
$\{2,4,1,3,5\}\ ,\ \{5,3,1,4,2\}\ ,\ \{4,3,1,5,2\}$
$\{5,4,1,3,2\}\hspace{31.6mm}\{2,4,1,5,3\}$
There are 8 so the answer is $8\cdot5=40$
A: I think the main fallacy in your argument is that you wrongly choose the properties
Let us take $p_1$ be the property that exactly one person has the same companion.
Number of ways in which this can be done is $\binom51*\binom21*3!$ Here there are $\binom21$ to choose companion(because in a round table each person has two companions and there are two ways to choose a companion who remains intact with the person.
Let us take $p_2$ be the property that exactly two person has the same companion.
It can be divided into cases 
i)When the two persons are at adjacent places.Number of ways to choose the persons=5(if the round table goes as $A->B->C->D->E->A$ Then it can be clearly seen that two adjacent persons can be choosen in 5 ways.)
hence number of ways=$5*2!$(we have only one option for choosing companion in this case)
2)When the two persons are one place next to each other 
Number of ways to choose the persons=$5$(Similar to argument in first case)
hence number of ways=$5*\binom32*2!$ Here there are $\binom32$ ways to choose companion(For example if round table is $A->B->C->D->E->A$ Then if $A$ and $C$ are choosen then possible choice of companions are $B,D,E$ of which any two is to be choosen.)One more thing here should be taken into consideration.When the two persons has same companion.It can be done in $5*2!$
3)Last case when the two persons are diagonally opposite to each other
CAUTION It is already counted in the second case.
Similarly when three persons has same companion=$2$(two rotations - clockwise or anticlockwise are possible)
Now apply IEP(which has none of the above properties) as $$4!-(\binom51*\binom21*3!)+(5*2!+5*\binom32*2!+5*2!)-2$$
