Generating function with tough restrictions 
In how many ways can a coin be flipped $25$ times in a row so that exactly $5$ heads occur and no more than $7$ tails occur consecutively?

For the heads, I think that it is $\binom{25}{5}$, but I do not know what to do with the tails restriction. 
Not sure how to approach this problem. It is from the advanced section in the generating functions chapter in my book.

More generally, we might ask how many sequences of flips of length $n$ have $m$ heads and no more than seven tails in a row.  We will find a bivariate generating function for this problem.
First suppose we drop the constraint that the sequence must contain exactly $m$ heads, retaining the requirement that there be no more than seven tails in a row; let's say such a sequence is "acceptable".  Let $f(z)$ be the generating function for the number of acceptable sequences of length $n$. An acceptable sequence consists of a sequence of zero to seven tails, or a sequence of zero to seven tails followed by a head, followed by an acceptable sequence. In terms of the generating function,this translates to
$$f(z) = (1+z+z^2+ \dots +z^7)  (1 + z f(z))$$
We can insert a "marker" for the head in this equation and convert the generating function into a bivariate generating function where the number of acceptable strings of length $n$ with $m$ heads is the coefficient of $u^m z^n$:
$$f(u,z) = (1+z+z^2+ \dots +z^7)  (1 + uz f(u,z))$$
Noting that $1+z+z^2+ \dots +z^7 = (1-z^8)/(1-z)$, we can solve the preceding equation for $f(u,z)$:
$$f(u,z) = \frac{(1-z^8)/(1-z)}{1-uz(1-z^8)/(1-z)}$$
In a sense this bivariate generating function is the general solution, but let's see if we can find the answer to the original problem, which is the coefficient of $u^5 z^{25}$ in $f(u,z)$.  We start by expanding $f(u,z)$ as a series in $u$.
$$f(u,z) = \frac{1-z^8}{1-z} \sum_{i=0}^{\infty} z^i \left(  \frac{1-z^8}{1-z} \right)^i u^i$$
The coefficient of $u^5$ is
$$\begin{align}
[u^5] f(u,z) &= \frac{1-z^8}{1-z}  z^5 \left(  \frac{1-z^8}{1-z} \right)^5  \\
&= z^5 (1-z^8)^6 (1-z)^{-6} \\
&= z^5 (1 - 6z^8 + 15 z^{16} - O(z^{24})) \sum_{i=0}^{\infty} \binom{6+i-1}{i} z^i
\end{align}$$
From this last equation we see that the coefficient of $u^5 z^{25}$ is
$$[z^{25}][u^5]f(u,z) = \binom{6+20-1}{20} -6 \binom{6+12-1}{12} + 15 \binom{6+4-1}{4} = \boxed{17,892}$$
