What is the probability that no letter is in its proper envelope? Five letters are addressed to five different persons and the corresponding
envelopes are prepared. The letters are put into the envelopes at random.
What is the probability that no letter is in its proper envelope?

Let $1,2,3,4,5$ be the letters and let $A,B,C,D,E$ be the proper envelope resepctively.
1) In case of five proper envelopes, only $1$ pattern.
2) In case of only four proper envelopes, the last one is in the proper envelope, so $0$ pattern.
3) In case of only three proper envelopes, you have $\binom{5}{3}\times 1=10$ patterns.
4) In case of only two proper envelopes, you have $\binom{5}{2}\times 2=20$ patterns.
For example, for $A=1, B=2,C,D,E$, you have two patterns as $(C,D,E)=(4,5,3),(5,3,4).$
5) In case of only one proper envelope, you have $\binom{5}{1}\times 9=45$ patterns.
For example, for $A,B,C,D,E=5$ you have nine patterns as
$$(A,B,C,D)=(2,1,4,3),(2,3,4,1),(2,4,1,3),(3,1,4,2),$$$$(3,4,1,2),(3,4,2,1),(4,1,2,3),(4,3,1,2),(4,3,2,1).$$
Hence, what you want is
$$1-\frac{1+0+10+20+45}{5!}=\frac{5!-76}{5!}=\frac{44}{120}=\frac{11}{30}.$$
