Answer: 840
Solution: Consider the result of the first throw. If it is $A$, since $A B$ appears three times, the sequence form should be $A \cdots B \cdots A \cdots B \cdots A \cdots B \cdots A \cdots$ $A$ or $A \cdots B \cdots A \cdots B \cdots A \cdots B \cdots B, B A$ appears at most three times, which contradicts the fact that $B A$ appears four times.

Therefore, the sequence form must be $B \cdots A \cdots B \cdots A \cdots B \cdots A \cdots B \cdots A \cdots$.
For $A A$, there are four positions, equivalent to the non-negative integer solutions of $x 1+x 2+x 3+x 4=2$. This is $C_{5}^{3}=10$;
For $\mathrm{BB}$, there are four positions, equivalent to the non-negative integer solutions of $x 1+x 2+x 3+x 4=6$, which is $C_{9}^{3}=84$;
In total, there are $10 \times 84=840$ sequences that meet the requirements.
6. Let the result of a single coin toss be denoted as $A$ if it lands heads up; as $B$ if it lands tails up. When tossing the coin several times in a row, we can obtain a sequence composed of $A$ and $B$, for example, tossing six times might result in the sequence $A A B A B B$. In the resulting sequence, we can count the occurrences of patterns like $A B$. For instance, in $A A B A B B$, $A B$ appears twice. When tossing the coin 16 times in a row, among all possible sequences, there are $\qquad$ sequences that satisfy: $A A$ appears 2 times, $B B$ appears 6 times, $A B$ appears 4 times, and $B A$ appears 4 times.