Permutation (without repeating)- (basic) I have problems solving and understanding the following task from the combinatorics:
We have two sets:
$\mathcal A=${$x; (x\in Z)$ $\land$ $(-6 \le x \lt 0)$} and $\mathcal B=${$n; (n\in N) \land (n|12)$}.
Let $ \mathcal G $ be the set of bijective mappings from the set $A$ to the set $B$.
So, according to the instructions of the task we can say that there's $6!= 720$ elements in the set $ \mathcal G $, right?
The questions is: For how many functions from the set $ \mathcal G $ for at least one element $x\in A$ counts $f(x)= -x$?
I don't know where or how to start. I would be really happy if somebody would try to explain to me this task. I'm having a test on this soon, and this is one of many tasks I need to know and understand how to solve- I'm really starting to worry about it.
Thank you in advance!
 A: Enumerate the complementary set $\mathcal{G}^c$:  bijective functions $f:\mathcal{A} \to \mathcal{B}$ such that $f(x) \ne -x$ for all $x$.  The five restrictions are that
$$
\left\{
\begin{align}
f(-6) &\ne 6 \\
f(-4) &\ne 4 \\
f(-3) &\ne 3 \\
f(-2) &\ne 2 \\
f(-1) &\ne 1.
\end{align}
\right.
$$
Notice that there is not an explicit restriction on the value $f(5)$.
Here's a nice way to recast the problem in terms of derangements.  Factor the map $f$ by writing $f = \beta \circ \pi \circ \alpha$, where $\pi$ is a permutation of the set $\{1, 2, 3, 4, 5, 6\}$ and $\alpha$ and $\beta$ are the explicit bijections pictured here:
$$
\begin{bmatrix} -1 \\ -2 \\ -3 \\ -4 \\ -5 \\ -6 \end{bmatrix}
\overset{\alpha}{\mapsto} \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ 6 \end{bmatrix}
\qquad \text{and} \qquad
\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ 6 \end{bmatrix}
\overset{\beta}{\mapsto} \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 12 \\ 6 \end{bmatrix}
$$
Since $\alpha$ and $\beta$ are bijections, the task of enumerating $\mathcal{G}^c$ amounts to "cleaner" task of enumerating permutations $\pi$ that don't fix any number except possibly $5$.  So, there are two cases.
Case $\pi(5) \ne 5$.  These are just derangements of $\{1, 2, 3, 4, 5, 6\}$.
Case $\pi(5) = 5$.  These are just derangements of $\{1, 2, 3, 4, 6\}$.
Therefore,
$$
\left| \mathcal{G}^c \right| = \; !6 \; + \; !5 = 265 + 44 = 309
$$
and
$$
\left| \mathcal{G} \right| = 720 - 309 = 411.
$$
