Combinatorics with postman and recepients A postman has to deliver five letters to five different houses. Mischievously, he posts one letter through each door without looking to see if it is the correct address. In how many different ways could he do this so that exactly two of the five houses receive the correct letters?
Please tell me why I'm wrong:
To give correct letters there one way for two houses, and to give wrong letters to remaining $3$, there are $3$ for $3^{rd}$, $2$ for $4^{th}$ & $1$ for last, so total is 6, but this is wrong, please help.

There are $\binom52=10$ different ways to choose the two houses that get the correct letters. Suppose that the other three houses are $A,B$, and $C$, and their correct letters are $a,b$, and $c$, respectively. What are the ways in which he can distribute the letters $a,b$, and $c$ so that no house gets the right letter? The problem is so small that the easiest approach is brute force: we’ll just list the ways. House $A$ has to get letter $b$ or letter $c$. If it gets letter $b$, the remaining letters are $a$ and $c$, and they have to go to houses $B$ and $C$. House $C$ can’t get letter $c$, so the only possibility is that house $B$ gets letter $c$, and house $C$ gets letter $a$. If house $A$ gets letter $c$, similar reasoning shows that house $B$ must get letter $a$ (since it can’t get letter $b$), and that leaves letter $b$ for house $C$. These two outcomes are shown in the table below:
$$\begin{array}{ccc}
A&B&C\\ \hline
b&c&a\\
c&a&b
\end{array}$$
Thus, for each of the $10$ ways to pick two houses to get the right letters there are $2$ ways for the other three houses all to get the wrong letter, so there are altogether $10\cdot 2=20$ different ways for the postman to deliver exactly two of the letters correctly.
