A question about permutations without fixed point I need to find how many permutations $f:[n]\to [n]$ exist that answer the following conditions:

  
*
  
*For every $i$: $f(i) \neq i$
  
*$ f(1)=2, f(2)=3$

Notes:
Here $[n] = \{x| 1 \le x \le n\}$.
Moreover, the number of permutations without a fixed point is: 
$$ n! \sum_{k=0}^n \frac{(-1)^k}{k!}$$
The answer has to do with the Inclusion–exclusion principle.
 A: We will use $D_n$ for the number of derangements on $n$ elements, that is,
$$D_n = \sum_{k = 0}^{n} (-1)^k\binom{n}{k}(n - k)! = n!\sum_{k = 0}^{n} \frac{(-1)^k}{n!}$$
We consider two cases.
Case 1: $f(3) = 1$
If $f(3) = 1$, we are left with a derangement of the remaining $n - 3$ elements, so there are $D_{n - 3}$ such cases.
Case 2: $f(3) \neq 1$.  
In this case, there are $n - 3$ ways to choose $j$ such that $f(j) = 1$.  
If $f(3) = j$, we are left with a derangement on $n - 4$ elements.  There are $(n - 3)D_4$ such cases. 
If $f(3) \neq j$, then we have $n - 3$ elements to assign, each with one forbidden place since $f(3) \neq j$ and $f(i) \neq i$ for $i \neq 3, j$.  Hence, we have a derangement on $n - 3$ elements.  Thus, there are $(n - 3)D_{n - 3}$ such cases.
Hence, there are $(n - 3)(D_{n - 4} + D_{n - 3})$ such cases.
Total:  Since the cases considered above are exhaustive and mutually exclusive, the number of permutations $f:[n] \to [n]$ such that $f(i) \neq i$ for all $i$, $f(1) = 2$, and $f(2) = 3$ is 
$$D_{n - 3} + (n - 3)(D_{n - 4} + D_{n - 3}) = (n - 2)D_{n - 3} + (n - 3)D_{n - 4}$$  
Note:  The argument given above is based on the derivation of the recursive formula 
\begin{align*}
D_0 & = 1\\
D_1 & = 0\\
D_n & = (n - 1)(D_{n - 1} + D_{n - 2}), n \geq 2
\end{align*}
for the number of derangements of a set with $n$ elements.
A: Denote by ${\cal F}_n$ the set of permutations $f:\>[n]\to[n]$ described in the question. To each such $f$ associate the derangement $g$ of $A_{n-1}:=[n]\setminus\{2\}$ defined by $$g(1)=3, \qquad  g(k)=f(k)\quad (3\leq k\leq n)\ .$$
Conversely, if a derangement $g$ of $A_{n-1}$ with $g(1)=3$ is given we obtain an $f$ described in the question by putting $$f(1):=2,\quad f(2):=3,\qquad f(k):=g(k)\quad(3\leq k\leq n)\ .$$ 
There are $D_{n-1}$ derangements of $A_{n-1}$, and ${1\over n-2}D_{n-1}$ of them have $g(1)=3$. It follows that
$$|{\cal F}_n|={1\over n-2} D_{n-1}\qquad (n\geq3)\ ,$$
which is equivalent to N.F. Taussig's answer.
