Probability of sending wrong email to all 5 people Say I have to write 5 emails to 5 people: Al, Ben, Charlie, Dan, Eli. Each email starts with the person's name (i.e., "Dear Al"). 
Mistakenly, as it was late at night, I wrote the To section to the wrong person (example below). 
Question: what is the probability that all emails will reach to the wrong person? 
Is this type of questions relates to the Inclusion–exclusion principle? If so, how would you solve it?

 A: Yes, one would need to use the inclusion-exclusion principle. The are 5! ways of sending the emails (universal set). There are |A| = 4! ways on which the 1st person receives the right mail, |B| = 4! ways in which the second person does, |C| = 4! ways in which the third person does ...... 
By the inclusion exclusion principle the number of ways in which at least one of them gets the right mail is [5(4!)] - [10(3!)] + [10(2!)] - [5] + [1] =  76. Therefore, the number of ways in which all of them get the wrong mail is 5! - 76 = 44.
So the probability would be 44/120 = 11/30
A: Permutations in which every person gets the wrong email are called derangements, which you can read about on that wiki page. The answer is the number of derangements of $5$, $!5 = 44$, divided by the number of permutations $5! = 120$. So
$$
\frac{!5}{5!} = \frac{44}{120} = \frac{11}{30}.
$$
For more detail you should refer to counting derangements, but you can use inclusion-exclusion to derive
$$
!n = n! \sum_{i=0}^n \frac{(-1)^i}{i!};
$$
in particular
\begin{align*}
!5
&= 5! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right) \\
&= \frac{120}{2} - \frac{120}{6} + \frac{120}{24} - \frac{120}{120} \\
&= 60 - 20 + 5 - 1 \\
&= 44.
\end{align*}
Also there is an elegant formula for $n \ge 1$,
$$
!n = \left\lfloor{\frac{n!}{e} + \frac12}\right\rfloor
$$
So you could also have gotten the answer by taking $120/e \approx 44.1455329406$, rounding to $44$, and dividing by $120$.
SEE ALSO: 1, 2, 3, 4,  on mathSE.
A: Here is a solution without using the Exclusion-Inclusion principe:
There are only two kinds ways everyone ends up with the wrong email:  Either you get a cycle of 5 (e.g. The emails to ABCDE get sent to BCDEA respectively), or you have a cycle of 3 plus a cycle of 2 (e.g. ABC get sent to BCA and DE to ED)
How many 5 cycles are there? Starting with A, the next is one of 4, the third of 3, etc. ... So that is 4! =24
How many 3+2 cycles? There are ${{5}\choose{3}}=10$ ways to divide up the 5 into 3 and 2, and the 3 cycle can be done in 2 ways, so that's 20.
Total: 24+20 = 44 ways for all of them to get a wrong email.
So, probability is $\frac{44}{5!}= \frac{44}{120}=\frac{11}{30}$
