1.52 In coin tossing, if $Z$ represents heads and $F$ represents tails, then the sequence of coin tosses is represented by a sequence of $Z$ and $F$. We can count the occurrences of heads followed by tails $(F Z)$, heads followed by heads $(Z Z)$, etc. For example, the sequence
ZZFFZZZZFZZFFFF

is the result of 15 coin tosses, which includes 5 $Z Z$, 3 $Z F$, 2 $F Z$, and 4 $F F$. How many sequences of 15 coin tosses have exactly 2 $Z Z$, 3 $Z F$, 4 $F Z$, and 5 $F F$?
[Solution] Clearly, a sequence of coin tosses can be regarded as being composed of alternating $Z$ segments and $F$ segments. When transitioning from a $Z$ segment to an $F$ segment, a $Z F$ appears at the boundary; when transitioning from an $F$ segment to a $Z$ segment, a $F Z$ appears.

Given that the sequence contains 4 $F Z$ and 3 $Z F$, the sequence must start with $F$ and end with $Z$, and is divided into 8 segments:
$$
(F)(Z)(F)(Z)(F)(Z)(F)(Z),
$$

where $(F)$ and $(Z)$ represent $F$ segments and $Z$ segments, respectively.
It is evident that if a segment of the same letter contains $k>1$ letters, it will produce $k-1$ $Z Z$ or $F F$. Since the sequence has exactly 2 $Z Z$, there must be a total of 6 $Z$ in the sequence. Among the 4 $Z$ segments, each segment is initially filled with 1 $Z$, and then the remaining 2 $Z$ are placed into 1 or 2 $Z$ segments, resulting in a total of $C_{4}^{1}+C_{4}^{2}=C_{5}^{2}=10$ different arrangements.

Since the sequence has 15 terms, and 6 of them are $Z$, the remaining 9 are $F$. Dividing 9 $F$ into 4 segments, let the number of $F$ in each segment be $x_{1}, x_{2}, x_{3}, x_{4}$, then we have
$$
x_{1}+x_{2}+x_{3}+x_{4}=9,
$$

The problem is reduced to finding the number of positive integer solutions to this equation. Clearly, there are $C_{8}^{3}=56$ solutions.
In summary, there are 560 sequences that meet the requirements of the problem.