Finding the number of 7 letter words from A, B, C, D, E, F, G in which A, B, C each occurs at least once.

Question

I've attempted doing this question using the inclusion-exclusion principle. But I have an answer which apparently isn't right. So I think my understanding might be a bit off. If someone could clarify where I went wrong or show me an alternate solution I'd appreciate it.

Finding the number of 7 letter words from A, B, C, D, E, F, G in which A, B, C each occurs at least once.

Total words = $7^7$

Words which don't include A, B, C = $3(6^7) - 3(5^7) - 4^7$

Answer 1

Let $X$ be the set of all seven letter words using these letters with no further restrictions. Let $N_A$ be the set of such words which contain no $A$'s. Let $N_B$ be the set of words which contain no $B$'s and $N_C$ no $C$'s.

You are trying to calculate $|N_A^c\cap N_B^c\cap N_C^c|$, the number of words containing at least one $A$ and at least one $B$ and at least one $C$.

We recognize that this is $|X\setminus (N_A\cup N_B\cup N_C)|=|X|-|N_A\cup N_B\cup N_C|$ which expands further by inclusion-exclusion principle as:

$$|X|-|N_A|-|N_B|-|N_C|+|N_A\cap N_B|+|N_A\cap N_C|+|N_B\cap N_C|-|N_A\cap N_B\cap N_C|$$

Each term of this should be familiar how to calculate at this point.

$7^7 - 3\cdot 6^7 + 3\cdot 5^7 - 4^7$

Note in particular how the signs change when expanding via inclusion-exclusion.