Number of Sequence Find the number of sequences of length $n$ consisting of the letters $a,b,c$ in which each letter occurs at least once.
My Approach: select 3 positions from $n$ positions and arrange the letters ${n\choose 3} 3!$ ways and arrange the remaining places in $3^{n-3}$ ways, resulting in a total of $3^{n-3}{n\choose 3}3!$ sequences.
Why is my approach giving me the wrong answer?

Let $X$ be the set of all words in $\{a,b,c\}$ of length $n$, and let $X_a,X_b,X_c$ be the set of all words in $X$ that do not contain the letter $a,b,c$, respectively.
Then you want to find $|X\setminus(X_a\cup X_b\cup X_c)|$.
Inclusion-exclusion to the rescue:
$$\begin{align}|X\setminus(X_a\cup X_b\cup X_c)|=&|X|\\
&-(|X_a|+|X_b|+|X_c|)\\
&+(|X_a\cap X_b|+|X_a\cap X_c|+|X_b\cap X_c|)\\
&-|X_a\cup X_b\cup X_c|
\end{align}$$
Each of these values is easier to compute.

The generating function approach is to note that the value we seek, $f(N)$, satisfies:
$$\sum_{n=0}^{\infty} \frac{f(n)}{n!}x^n = (e^x-1)^3=e^{3x}-3e^{2x}+3e^x-1$$
and expand the right side out.
The generating function approach is nice for alternate questions. If there is no condition on $a$, and at least one $b$ and two $c$s, then you'd get the generating function:
$$e^x(e^x-1)(e^x-1-x)=e^{3x}-e^{2x}(x+2)+e^{x}(x+1)$$
So the coefficient of $x^n$, when $n>0$, is $$\frac{3^n}{n!} - \frac{2^{n+1}}{n!}-\frac{2^{n-1}}{(n-1)!}+\frac{1}{n!}+\frac{1}{(n-1)!}$$
So the count is $n!$ times this, or:
$$3^n-2^{n+1}-n2^{n-1}+1+n=3^n-2^{n-1}(4+n)+(n+1)$$
