Given $7$ blank spaces and four letters $A,B,C,D$, what is the number of ways each space can be filled up so that all four letters are used up. Given $7$ blank spaces and four letters $A,B,C,D$, what is the number of ways each space can be filled up so that all four letters are used up.
The only way I can think of is: Fill up all the squares in any way possible, then remove the arrangements that use less than $4$ letters. This can be done in $4^7 - 4 \times3^7$.
Is this correct? Is there a way we can use PIE (Principle of Inclusion-Exclusion) to solve this?

This question is very good place to use exponential generating functions .
It is said that all letter will be used  ,so
the generating function for $A$ is $$x + \frac{x^2}{2} +\frac{x^3}{6} + \frac{x^4}{24}$$
the generating function for $B$ is $$x + \frac{x^2}{2} +\frac{x^3}{6} + \frac{x^4}{24}$$
the generating function for $C$ is $$x + \frac{x^2}{2} +\frac{x^3}{6} + \frac{x^4}{24}$$
the generating function for $D$ is $$x + \frac{x^2}{2} +\frac{x^3}{6} + \frac{x^4}{24}$$
We wrote our exponential generating functions starting from $x$ because each letter must be used and if each letter will be used then a letter can be used at most $4$ times.
Now, we should find their product . Moreover ,  when we find their expansion , we should look for the term $[x^{7}]$ and multiply it by $7!$.
Such that https://www.wolframalpha.com/input/?i=expanded+form+of+%28x+%2B+x%5E2+%2F+2+%2B+x%5E3+%2F+6++%2B+x%5E4+%2F+24%29%5E4
We found that the coefficient of $[x^7]$ is $\frac{5}{3}$ ,so $7! \times \frac{5}{3}=8400$
