Counting the number of surjections.

You’ve actually made a pretty good start: your first idea doesn’t work, but it’s a reasonable one to try, and you’ve spotted the difficulty with it. What you want now is an inclusion-exclusion argument. There are $3^n$ functions altogether. $2^n$ of them are functions from $\{1,\dots,n\}$ to $\{A,B\}$, missing $C$ altogether, so we need to throw them out. There are also $2^n$ functions that miss $B$ and $2^n$ that miss $A$, so our improved approximation to the number of surjections is $3^n-3\cdot2^n$.

Unfortunately, we’ve now gone overboard. The function that sends every $k\in\{1,\dots,n\}$ to $A$, for instance, has been thrown out twice, once for not hitting $B$ and once for not hitting $C$. The same goes for the other two constant functions. We need to throw each of them back in once, so that on net we’ve thrown each of them out just once, not twice. The revised number of surjections is then

$$3^n-3\cdot2^n+3=3\left(3^{n-1}-2^n+1\right)\;.\tag{1}$$

A little thought should convince you that no further adjustments are required and that $(1)$ is therefore the desired number.

Ignoring surjectivity, there are $3^n$ maps $\{1,2,\ldots,n\}\to \{A,B,C\}$. We should subtract the $2^n$ maps that ar actually just maps $\{1,2,\ldots,n\}\to \{A,B\}$, also the $2^n$ maps $\ldots\to \{A,C\}$ and the $2^n$ maps $\ldots\to \{B,C\}$. But now we have subtracted too much: The three constant maps $f(x)=A$, $f(x)=B$ and $f(x)=C$ where subtracted twice (e.g. $f(x)=A$ as a map $\to\{A,B\}$ and also as a map $\{A,C\}$). In summary there are therefore $$ 3^n-3\cdot 2^n+3$$ surjective maps $\{1,2,\ldots,n\}\to\{A,B,C\}$.

Using PIE one can find a more generalized form of the number of surjections. The number of surjections of a set containing $m$ elements onto a set containing $n$ elements ($m>n$) is

$$\sum\limits_{i=0}^{n}(-1)^{i}\dbinom{n}{i}(n-i)^{m}$$

In the given question $n=3$ and $m=n$.