Derangement related problem-bijective functions $f: A \to A$ such that $f(x) \neq x$ and $f(1) \neq 2$ Let $R$ denote the set of all nonempty relations on the set $A$, where $A = \{1, 2, 3, 4, 5\}$. A function $f(x)$ is chosen at random from set $R$. The probability that $f(x)$ is bijective, $f(x)\ne x$ such that $x\in A$ and $f(1)\ne 2$ is_____
My approach is as follow total number of relations is $5 \cdot 5=25$.
Hence, number of nonempty relations is $2^{25}-1$.
Given the formula of derangement where $f(x)\ne x$ we get $5! \cdot (1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!})=44$, I am not able to insert condition $f(1)\ne 2$ because those case needs to be removed where $f(x)=2$.
 A: Let us find the number of derangements $\left\{ 1,2,3,4,5\right\} \to\left\{ 1,2,3,4,5\right\} $
that satisfy $f\left(1\right)=2$.
This equals the number of bijections $\left\{ 2,3,4,5\right\} \to\left\{ 1,3,4,5\right\} $
that satisfy $f\left(k\right)\neq k$ for $k\in\left\{ 3,4,5\right\} $.
Let $A_{k}$ denote the set of bijections with $f\left(k\right)=k$.
Then applying the principle of inclusion/exclusion and also symmetry we find: $$\left|A_{3}^{\complement}\cap A_{4}^{\complement}\cap A_{5}^{\complement}\right|=4!-\left|A_{3}\cup A_{4}\cup A_{5}\right|=4!-3\left|A_{3}\right|+3\left|A_{3}\cap A_{4}\right|-\left|A_{3}\cap A_{4}\cap A_{5}\right|=$$$$24-3\times3!+3\times2!-1!=11$$
So if there are indeed $44$ derangements in total then $44-11=33$ of them will satisfy $f(1)\neq2$.

addendum
After a second look I realized that things can be solved a lot easyer (without using PIE).
The set of derangements $\left\{ 1,2,3,4,5\right\} \to\left\{ 1,2,3,4,5\right\} $ can be split up in $4$ disjoint subsets: $D_2,D_3,D_4,D_5$. Here $D_i$ denotes the set of derangements that satisfy $f(1)=i$. By symmetry it is clear that the sets have equal cardinality, so if the sum of these cardinalities equals $44$ then the cardinality of $D_3\cup D_4\cup D_5$ is $33$.
