Permutations of $[5]$ such that $\sigma (4)\ne 2$ and $\sigma (1)\notin \{1,2\}$?

Question

How many permutations of $[5]$ are there such that $\sigma (4)\ne 2$ and $\sigma (1)\notin \{1,2\}$?

My approach: Define sets $A$ and $B$ as

$A=\{\sigma:\sigma(4)=2\}$

$B=\{\sigma:\sigma(1)\in \{1,2\}\}$

So, our answer will be $5!-(\color{red}{|A|}+\color{blue}{|B|}-\color{green}{|A\cap B|})=5!-(\color{red}{4!}+\color{blue}{2\cdot4!}-\color{green}{3!})=\fbox{54}$

Is this correct?

Answer 1

Yes, this looks correct. If you would prefer to avoid the inclusion-exclusion you can choose first $\sigma(1)$, then $\sigma(4)$ and then the rest to get $3 \times 3 \times 3! = 54$.