Example 9 In a sequence of coin tosses, the number of times a tail is followed by a head (denoted as "tail-head"), a head is followed by a tail (denoted as "head-tail"), a head is followed by a head (denoted as "head-head"), and a tail is followed by a tail (denoted as "tail-tail") can be counted. How many different sequences of 15 coin tosses contain 2 "head-head", 3 "head-tail", 5 "tail-tail", and 4 "tail-head"?
(4th AIME)
Analysis: From "反" being 1 more than "正", we know that the sequence starts and ends with "反" and "正" respectively. According to the conditions, it is easy to know that the count of "正" equals $2 \times 2 + 3 + 4 = 11$, and the count of "反" is $5 \times 2 + 3 + 4 = 17$. Since the "反" and "正" at the beginning and end are each counted once, while the other "正" and "反" are each counted twice, there are 6 "正" and 9 "反" in the sequence.
The above analysis is the key to solving this example.
Solution: Construct the sequence model as follows:
反0000000000000正 where the empty positions "○" are sequentially denoted as $a_{i}(i=1,2, \cdots, 13)$.
Notice that there are three possible cases for 2 "正正":
(1) If $a_{13}$ is "反", then the remaining 5 "正" can only be divided into $(2,2,1)$ or $(3,1,1)$. When the 5 "正" are divided into $(2,2,1)$, there are still 7 "反" in $a_{i}$ (where $1 \leqslant i \leqslant 12$), and the 5 "正" can be inserted into 3 of the 8 gaps between the 7 "反", like:
"○反○反正反○反○反○反○反○" hence there are $\mathrm{C}_{8}^{3} \mathrm{C}_{3}^{1}$ different sequences.
Similarly, when the 5 "正" are divided into $(3,1,1)$, there are also $\mathrm{C}_{8}^{3} \mathrm{C}_{3}^{1}$ different sequences.

Therefore, when $a_{13}$ is "反", there are $2 \mathrm{C}_{8}^{3} \mathrm{C}_{3}^{1}$ different sequences.
(2) If $a_{13}$ is "正" and $a_{12}$ is "反", then the remaining 4 "正" can be divided into $(2,1,1)$, yielding $\mathrm{C}_{8}^{3} \mathrm{C}_{3}^{1}$ different sequences.
(3) If both $a_{13}$ and $a_{12}$ are "正", then the remaining 3 "正" can only be divided into $(1,1,1)$, and $a_{11}$ must be "反", resulting in $\mathrm{C}_{8}^{3}$ different sequences.
In summary, the total number of different sequences is
$$
2 \mathrm{C}_{8}^{3} \mathrm{C}_{3}^{1} + \mathrm{C}_{8}^{3} \mathrm{C}_{8}^{1} + \mathrm{C}_{8}^{3} = 560 \text{ (sequences). }
$$