Letters in Mailboxes so that none are in the right one

Question

In how many ways can 5 letters be put in 5 mailboxes such that none are placed in the right one ? I creatively thought labeling the letters & mailboxes 1-5 , it would be saying letter 1 can be placed in any of the other four . So a total of 4^5 ways .this was not one of the answers . Then i thought hmm lets write as an example 13254 as a sequence above 12345 ..this sequence is a ' throw-out ' case because the 1 cant go with the 1. This would generate _ _ a two digit number ..one for the letter and one for the mailbox .as an example 12 is legit . ( letter 1 in mailbox 2 ) while the pairs 11,22, ect are not . This would give 5 possabilities for the first number and 5 for the second , or 5x5 =25. But then we must subtract out the posabilities of the 11, 22, 33, ect This gives 25- 5 =20 but that is STILL not one of the answers

Answer 1

These are known as derangements.

Look here: https://en.wikipedia.org/wiki/Derangement

Answer 2

So you count bijective functions from $\left\{1,2,3,4,5\right\}$ to $\left\{1,2,3,4,5\right\}$ such that $f(i) \ne i$

I would do like this: Let $A_i$ be a set of functions with $f(i)=i$. Then

$|A_i|= 4!$ and

$|A_i\cap A_j|=3!$ and

$|A_i\cap A_j\cap A_k|=2!$ and

$|A_i\cap A_j\cap A_k\cap A_n|=1$

You are interested in $|A_1'\cup A_2'...\cup A_5'|$. Let's use PIE

$$|A_1'\cup A_2'...\cup A_5'| = 5!-5\cdot 4!+{5\choose 2}3!-{5\choose 3}2!+ {5\choose 4}1!- {5\choose 5}\cdot 1=...$$