Basic question. Does not involve permutations/combinations. $3$ mailboxes containing $3$ letters My teacher explained this problem to us - "There are $3$ mailboxes. $3$ people put letters in at random. There is no preference for any of the $3$ mailboxes.  Compute the probability that each mailbox contains $1$ letter."
I tried this problem on my own and got the wrong answer. I understand the teacher's solution but I don't get what went wrong with my answer. Can someone explain where I went wrong?
My answer: Treat each outcome as a sequence of $3$ numbers denoting the number of letters in each mailbox, e.g $300$ means $3$ in the 1st, $0$ in the second and $0$ in the 3rd mailbox. 
Sample space $S = \{300, 030, 003, 111,012,021, 120,102,210,201\}$
$S$ contains $10$ sample points.
Let $A =$ event that each mailbox is chosen once.
$P(A) = 1/10$ 
Teacher's solution: Treat each outcome as a sequence of $3$ numbers denoting the person and their corresponding chosen mailbox, e.g $312$ means the first person chose mailbox 3, the second chose mailbox 1, the third chose mailbox 3. So the sample space is $S = \{ 111, 112...\}$ and contains $27$ sample points.
Now let $A =$ event that each mailbox is chosen once.
$A =\{123,132,213,231,312,321\}$
So the answer is $P(A) = 6/27$
 A: The problem with your approach is that not every one of those outcomes have the same probability. In your sample space the probability of $300$ is not $\frac{1}{10}$, it is $\frac{1}{27}$. This is because there is no preference, you can multiply the probabilities of each event individually. The first letter has $\frac{1}{3}$ just like the second and the third. Another way of looking at it is that there are multiple ways to get to $111$ but only one way to get to $300$.
A: Please note that each letter is different so $111$ can have $3!$ ways same for $1,2$ so there are more than $10$ ways . thus this is the flaw of your solution put a,b,c as letters and A,B,C as mailboxes and then count ways but still personally i think teachers solution is fast and easy as always writing sample space isnt useful.
A: By counting how many envelopes were placed in each mailbox, you treated the three letters as if they were identical.  However, they are distinct.  Consequently, the outcomes you listed are not equally likely.
Suppose the letters are $a$, $b$, and $c$ and the mailboxes are $A$, $B$, $C$.  Since there are three mailboxes in which each letter can be placed, there are $3^3 = 27$ equally likely assignments of letters to mailboxes.  They can be listed as ordered pairs.  For instance, $(a, C)$, $(b,A)$, $(c,A)$ is the assignment of the first letter to the third mailbox and the other two letters to the first mailbox.
There is exactly one way to place all three envelopes in the first mailbox, namely by making the assignment $(a, A)$, $(b, A)$, and $(c, A)$.  There is also exactly one way to place all three envelopes in the second mailbox and exactly one way to place all three envelopes in the third mailbox.
On the other hand, there are $3! = 6$ ways to place the three letters in three different mailboxes. For such an assignment to occur, once the first person to arrive has selected one of the three mailboxes, the next person must select one of the other two mailboxes, and the final person must choose the mailbox the first two people did not select.  By counting how many envelopes were placed in each mailbox, you treated these six outcomes as identical.  
Similarly, by counting how many envelopes were placed in each mailbox rather than which envelopes were placed in each mailbox, you treated each of the six cases in which two letters were placed in one mailbox and one was placed in another mailbox as one case. Each such case can occur in $3$ ways since there are $3$ ways of selecting which letter will not be in the mailbox that contains two letters.
If $(k, m, n)$ denotes $k$ letters in mailbox $A$, $m$ letters in mailbox $B$, and $n$ letters in mailbox $C$, then the corresponding number of outcomes is 
\begin{array}{c c}
\text{case} & \text{number of outcomes}\\
(3, 0, 0) & 1\\
(2, 1, 0) & 3\\
(2, 0, 1) & 3\\
(1, 2, 0) & 3\\
(1, 1, 1) & 6\\
(1, 0, 2) & 3\\
(0, 3, 0) & 1\\
(0, 2, 1) & 3\\
(0, 1, 2) & 3\\
(0, 0, 3) & 1
\end{array}
As you can see, there are a total of $27$ outcomes. By counting how many envelopes were in each mailbox, you reduced the problem to $10$ cases that are not equally likely.
