A generalization to this pattern? $n$ persons has an object, they exchange their objects with others one time each one. How many forms can be found for the case in which each person ended with other object different from where they started with?
For example:


*

*1 person has 0 ways to end with other's object.

*2 persons have 1 ways to end with other's object.

*3 persons have 2 ways to end with other's object.

*4 persons have 12 ways to end with other's object.


What you're looking for is known as a Derangement of a permutation. Basically, given $n$ objects, you want to find out the number of ways to rearrange them so that no object is in its original position. Sometimes this number is denoted by $!n$, which definitely does not have the same value as $n$ factorial which is denoted $n!$. 
There are multiple ways to express recursive relationships and some closed-form results for counting derangements. By taking base cases as $!0 = 1$ and $!1=0$, a recurrence relation can be defined like so:
$$
!n = (n-1)[!(n-1) + !(n-2)].
$$
My favorite derangement formula in non-recursive forms is
$$
!n = n! \sum\limits_{k=0}^{n} \frac{(-1)^k}{i!}.$$
And this one, which is an elegant closed form:
$$
!n = \left[ \frac{n!}{e} \right]
$$
where the brackets indicate the nearest integer function.
