Let $n \ge$ 3 ($n \in \mathbb{N}$).
How many permutations $\sigma [n] \to [n]$ there are such that:
$\forall i: \sigma (i) \ne i$,
$\sigma (1) = 2$,
$\sigma (2) = 3$.
I know that I'm supposed to use inclusion-exclusion, and clearly it has something to do with derangements.
I have problem calculating the cardinalities of the sets I define. Can someone give me a guideline for the solution?
You are right we use derangement. Since f(1)=2 ; f(2) =3 For rest f(i) is not I. ;; Two case are possible CASE 1 :: f(3) =1 then number of permutations are $$ D_{(n-3)} $$ I.e Derangement of rest n-3 objects at (n-3) places ;;
Case 2 :: if f(3) is not 1 then $$ D_{(n-2)} $$
Total permutations are $$ D_{(n-3)} + D_{(n-2)} $$