Derangements of four objects

Let $A_i$ be the set of permutations with $\sigma(i)=i$. A permutation with no fixed point lies in none of the $A_i$. For any $j$ indices, the permutations fixing all of them number $(4-j)!$. By inclusion–exclusion, the number of derangements is
$$4!-\binom41\,3!+\binom42\,2!-\binom43\,1!+\binom44\,0!=24-24+12-4+1=9.$$