There is 9-ways to 4 couple to dance with each other, when nobody can dance with his/her own partner. The statement: There is 9-ways to 4 couple to dance with each other, when nobody can dance with his/her own partner.
Is true or not? And why? How can one calculate this? 
This means 4 girl and 4 boy, every boy can dance with 3 girl, but repetition should be subtracted, isn't it?
Thanks for the help!
Edit: Is this question the same?
Edit2.: The accepted answer solves my question how I wanted to solve it, Ross Millikan provided the simplest and most convenient solution.
 A: If you imagine lining the couples up, then you need to make a derangement of the men.  The linked article shows there are the closest integer to $\frac {4!}e$ of those, which is $9$.
A: I would suggest solving this problem by inclusion-exclusion.
Assuming that we can label the couples $1,2,3,4$... Let $A_1$ be the set of all pairings where couple $1$ is matched up, $A_2$ be the set of all pairings where couple $2$ is matched up, and so on. Then, you can count the number of elements in $A_i$, $A_i \cap A_j$, $A_i \cap A_j \cap A_k$, and $A_1 \cap A_2 \cap A_3 \cap A_4$. Once you do that, you can use inclusion-exclusion to find out how many elements are in $A_1 \cup A_2 \cup A_3 \cup A_4$, which is the set of all pairings in which there is at least one couple dancing together. The complement is the set of all pairings in which there is no couple dancing together.
A: You can walk through on cases - it's not too bad. We have partners $(a_i, b_i), i=1..4$ and allocating the $a_i$ :
$$ \begin{array}{c|c|c}
a_1 \to b_2 & a_2 \to b_1 & a_3 \to b_4 * & a_4 \to b_3 \\
a_1 \to b_2 & a_2 \to b_3 & a_3 \to b_4 * & a_4 \to b_1 \\
a_1 \to b_2 & a_2 \to b_4 & a_3 \to b_1 * & a_4 \to b_3 \\
 \end{array}$$
In each case the * indicates a forced choice to allow completion of the matching under the rules.
By symmetry on $a_1$'s allocation, there are $3\times 3 = 9$ possibilities.
