Let $n\ge3$ be an integer. How many permutations $f : [n]\to [n]$ are in which $f(i)\neq i$ for each $i\in \{2,3,...,n\}$ and $f(1)=1$, $f(2)=3$?

Question

Let $n\ge 3$ be an integer. How many permutations $f : [n]\to [n]$ are in which $f(i)\neq i$ to each $i \in \{2,3,...,n\}$ and $f(1)=1$, $f(2)=3$?

I need to use the Inclusion-exclusion principle but I don't know how...

Thank you for you help:)

Answer 1

Take $\phi = (2,3)$, the transposition that exchanges $2$ and $3$ and leaves all other points fixed. Then $f$ satisfies your properties if and only if $g := f \circ \phi$ leaves $1$ and $3$ fixed and nothing else, so we might as well count these $g$.
Reducing the problem even further, we have to count fixpoint-free permutations on $\{2,4,5,6,\ldots, n\}$. Now this is a well-known number, see for example here, and the result will be $$\left\lfloor \frac{(n-2)!}{e} + \frac{1}{2} \right\rfloor.$$

edit: Added the +1/2 missing in the linked number, both the Wikipedia page given by antkam in the comment and OEIS state that this is the right formula.

Answer 2

Because $f(1)=1$ we can say that this number equals the number of derangements $f:\{2,\dots,n\}\to\{2,\dots,n\}$ that satisfy $f(2)=3$.

If $D$ denotes the set of all derangements $f:\{2,\dots,n\}\to\{2,\dots,n\}$ then we can write $D=\bigcup_{k=3}^nD_k$ where $D_k$ is the set of elements that satisfy $f(2)=k$.

Evidently the $D_k$ are equinumerous and disjoint so that $|D|=(n-2)|D_3|$ hence in order to find $|D_3|$ it is enough to find $|D|$.

Can you do that yourself on base of the principle of inclusion/exclusion?

It comes to finding: $$(n-1)!-|A_2\cup\cdots\cup A_n|$$ where $A_i$ stands for the set of permutations on $\{2,3,\dots,n\}$ that satisfy $f(i)=i$.

---

Answer 3

There are two possibilities here: either $f(3) = 2$, in which case you must have a derangement on the remaining $n - 3$ elements, or $f(3) \neq 2$. If $f(3) \neq 2$, then each of the $n - 2$ elements other than $1$ and $2$ have one prohibited position, so you have a derangement on the other $n - 2$ elements. Hence, the number of admissible arrangements is simply $D_{n - 3} + D_{n - 2}$.

Sanity check: Note that $D_0 = 1$ since no element of the empty set is in its proper place and $D_1 = 0$ since the only permutation of $\{1\}$ is the identity. The number of permutations of $\{1, 2, 3\}$ such that $f(1) = 1$, $f(2) = 3$, and $f(3) \neq 3$ is $1 = D_0 + D_1$ since $3$ must be mapped to $2$.