We are tasked with calculating how many possible Aopslandian license plates can be created with exactly four $A$'s, exactly two $B$'s, or both. License plates consist of six uppercase letters, and no two license plates are the same.

Step 1: Defining Sets

Let:

• $S_A$ represent the set of license plates with exactly four $A$'s.
• $S_B$ represent the set of license plates with exactly two $B$'s.
• $S_{A \cap B}$ represent the set of license plates with exactly four $A$'s and exactly two $B$'s.

We need to find: $|S_A \cup S_B| = |S_A| + |S_B| - |S_{A \cap B}|$ where $|S_A|$, $|S_B|$, and $|S_{A \cap B}|$ are the sizes of the respective sets.

Step 2: Calculating $|S_A|$ (Plates with exactly 4 A's)

For $S_A$, we need exactly four $A$'s and the remaining two positions filled with letters other than $A$.

• Choose 4 positions out of 6 to place the $A$'s. The number of ways to do this is: $\binom{6}{4} = 15$
• The remaining 2 positions can be filled with any of the 25 other letters (since we exclude $A$). Thus, for each of these positions, there are 25 choices, giving: $25 \times 25 = 625$ Thus, the total number of license plates in $S_A$ is: $|S_A| = 15 \times 625 = 9375$

Step 3: Calculating $|S_B|$ (Plates with exactly 2 B's)

For $S_B$, we need exactly two $B$'s and the remaining four positions filled with letters other than $B$.

• Choose 2 positions out of 6 to place the $B$'s. The number of ways to do this is: $\binom{6}{2} = 15$
• The remaining 4 positions can be filled with any of the 25 other letters (since we exclude $B$). Thus, for each of these positions, there are 25 choices, giving: $25 \times 25 \times 25 \times 25 = 25^4 = 390625$ Thus, the total number of license plates in $S_B$ is: $|S_B| = 15 \times 390625 = 5859375$

Step 4: Calculating $|S_{A \cap B}|$ (Plates with exactly 4 A's and 2 B's)

For $S_{A \cap B}$, we need exactly four $A$'s and exactly two $B$'s.

• Choose 4 positions out of 6 to place the $A$'s. The remaining 2 positions must be filled with $B$'s. The number of ways to do this is: $\binom{6}{4} = \binom{6}{2} = 15$ Thus, the total number of license plates in $S_{A \cap B}$ is: $|S_{A \cap B}| = 15$

Step 5: Applying the Inclusion-Exclusion Principle

Now, using the inclusion-exclusion formula: $|S_A \cup S_B| = |S_A| + |S_B| - |S_{A \cap B}|$ Substitute the values we calculated: $|S_A \cup S_B| = 9375 + 5859375 - 15 = 5868735$

Thus, the total number of possible Aopslandian license plates containing exactly four $A$'s, exactly two $B$'s, or both is $\boxed{5868735}$.

Would you like a deeper breakdown of any steps or have further questions?

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Here are five related questions for practice:

1. How many license plates are there with exactly three $A$'s and three different letters?
2. What if license plates could also contain digits? How would the total count change?
3. How many license plates are there if they must contain exactly one vowel?
4. What if we wanted plates with exactly three of any single letter? How would that calculation differ?
5. How many license plates have no repeated letters at all?

Tip: In combinatorics, inclusion-exclusion is essential for handling problems with overlapping constraints.