First, arrange the 8 consonants (with one repeat of a double C) in $\frac{8!}{2!}$ distinct ways.
Now arrange the five vowels, with two repeats, double I, and double O, in $\frac{5!}{2! *2!}$ ways.
Finally, select 5 of the 9 slots for vowels, one at the start of the consonants, and one after each consonant.
Multiply. Done.
Let $W(c,n)$ denote the number of words of length $c$ from an alphabet of $n$ letters. Then $W(c,n)=n^c$.
Out of these, the number of words of the same size that do not contain one of the letters is $W(c,n-1)=(n-1)^c$. The number of ways of choosing which letter is missing is $\binom{n}{1}$.
The number of words of the same size that do not contain two letters is $W(c,n-2)=(n-2)^c$. The number of ways of choosing which two letters are missing is $\binom{n}{2}$... and so on ...
Now we use inclusion-exclusion principle: (subtract the number of words missing one of the letters, then add the number missing two of the letters, subtract the number missing three of the letters,...)
We get:
$$W(c,n)-\binom{n}{1}W(c,n-1)+\binom{n}{2}W(c,n-2)-\binom{n}{3}W(c,n-3)+\cdots+(-1)^{n-1}\binom{n}{n-1}W(c,n-(n-1)).$$
This is
$$n^c-\binom{n}{1}(n-1)^c+\binom{n}{2}(n-2)^c-\binom{n}{3}(n-3)^c+\cdots+(-1)^{n-1}\binom{n}{n-1}1^c.$$
or
$$\sum_{k=0}^{n-1}(-1)^k\binom{n}{k}(n-k)^c.$$
Another way could be: Denote $S_c^n$ the number of ways to partition the word of length $c$ into $n$ pieces. Then we just need to choose which letter goes to each of the $n$ pieces. This number is $n!$. So the number of words we are looking for is
$$n!S_c^n.$$
The numbers $S_c^n$ are called Stirling's numbers of the second kind.
Best Answer
The letter $z$ is always followed by a non-$z$, except for the $z$ at the end. So group each $z$ with letter after it to form a double letter. The string is built out of letters and double letters; call each of these a "unit". The generating function of a single unit is $25x+25x^2$; there are $25$ letters of length $1$, and $25$ double letters of length $2$. Since we want a sequence of these units, the g.f. is $$ {1\over 1-25x-25x^2} $$ But wait, we forgot about the possible $z$ at the end! There either is or is not a $z$ at the end, which is accounted for by a factor of $1+x$. Therefore, the answer is actually $$ \bbox[5px,border:1.5px solid black]{{1\over 1-25x-25x^2}\cdot (1+x).} $$