Drunken Secretary

The number of derangements of 4 is 9. There are 4! ways to send out the emails. So the answer is $\frac{9}{4!} \times 24 =9$

In the following solution I will use casework and complementary counting.....

$\large \ total \ number \ of \ choices \ = 4! = 24 \ (1)$

$\large \ Number \ with \ all \ correct \ = 1 \ (2)$

Now it is impossible to have 3 correctly sent letters because the 4th one will automatically be sent to another correct e-mail address. To find the number of options for two correctly sent email, we consider that the probability of a single email being correctly sent = $\frac{1}{4}$ . So: $\large \ Number \ with \ two \ correct \ = \frac{4!}{4} = 3! = 6 \ (3)$

For one letter correctly sent, we are left with 3 emails and email addresses. The probability of the 1st one being wrong is $\frac{1}{3}$ and the probability of the 2nd one being wrong is $\frac{1}{2}$ (the last one just has 1 choice, so it is ignored). So:

$\large \ Number \ with \ one \ correct = 4! \times\frac{2}{3}\times\frac{1}{2} = 8 \ (4)$

From equations 1,2,3 and 4: $\large \ number \ of \ (all)incorrect \ emails$ $= 24 - (6+8+1) = \boxed{9}$ $\therefore\ P(all \ wrong ) = \frac{9}{24} = \frac{3}{8}$

why you dont use this:

$i!\cdot \sum _{n=0}^4\left(\frac{\left(-1\right)^n}{\:n!}\right)$

which gives:

4! * $\frac{3}{8} = 9 \boxed{ANS}$

we subtract all the possibilities with 1 or more correctly sent emails. 4 correctly sent emails: 1

3 correctly sent emails: redundant because 3 correctly sent makes the last one trivially obvious that it reaches the right recepient

2 correctly sent emails: 4C2 = 6, if at least two are correctly sent, the remaining two has two options: gets correctly sent or wrongly sent, therefore only one combination for every possible 2 email combination that's chosen to be correct

1 correctly sent email: for the remaining 3 emails yet to be sent, we count out the possibility that another email or another 2 emails gets sent correctly after the first one. therefore, for the remaining 3, there's only two permutations to make the remaining 3 all wrong. Thus, 4 x 2 = 8 number of all possible permutations: 4! = 24

$\frac{24-8-6-1}{24}$