A combinatoric question using the Inclusion–exclusion principle I was given the next question: A secretary has to send 3 letters to person A and 2 letters to person B. The secretary prepared 5 envelopes with 3 of them named A and 2 of them named B.  but being the clumsy secretary she is, she put the letters at a random order. What is the number of possibilities that she sent all of them correctly?
I'd like some help with understanding where the inclusion-exclusion principle is implemented, and if not, there is another way of trying to solve this question? Thanks!
 A: This does not seem to require Inclusion-Exclusion as the number of ways to send the correct letters to A is $3!=6$ and the number of ways to send the correct letters to B is $2!=2$. Therefore, the number of ways to send the correct letters to both is $12$.
The number of ways to send the letters to both is $5!=120$. Thus, the probability that the letters were sent correctly is $\frac1{10}$.
A: The problem doesn’t state this, but I’m going to assume she sends the first three letters to A and the second two letters to B. That seems to be the direction the problem is pointing at.
Now, if A gets the correct letters, then B will automatically do so as well. So in our equivalent problem, all the letters are sent correctly if and only if the first three letters are sent to A.
The problem also fails to state how the letters are mixed up, so let’s assume uniformity at random. There are $5$ choice of letters for the first letter in order, and $3$ of them are correct, so there’s a $3/5$ chance that the first letter is sent to $A$. Now there are $4$ letters left and $2$ of them are correct, so there’s a $2/4=1/2$ chance that the second letter is also sent correctly to $A$. Finally, there’s a $1/3$ chance that the third letter is correctly sent to $A$. Multiplying these all together gives a probability of $1/10$. There are $5!$ total orderings and $1/10$ of them are correct, so the number of possible orderings that are correct is $5!/10=12$.
For a second approach, put the letters in a correct ordering. We now wish to count the number of swaps we can do such that the letters all still go to the right people. That means we can perm the the first three letters in any order, which gives $3!=6$ orderings, and we can also permute the last two letters is any order, which gives $2!=2$ orderings. Multiplying these together again yields the answer: $12$
