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We'll look separately at the cases where we have $0, 1, 2$ and $3$ non-consecutive $x$ 's.

(o) no $x$ 's: then there are $2$ choices for each of the $5$ positions in the word, giving us $2^{5} = 32$ words;

(i) one $x$ : start by placing the $x$ in one of the $5$ positions, then fill the remaining positions in $2^{4} = 16$ ways, giving us $5*16 = 80$ more words;

(ii) two $x$ 's: there are $6$ possible position pairs for the two $x$ 's, namely $(1,3), (1,4), (1,5), (2,4), (2,5), (3,5)$ . The remaining two positions can then be filled in $2^{3} = 8$ possible ways, giving us $48$ more words;

(iii) three $x$ 's: the $x$ 's can only go in positions $1,3$ and $5$ , with the remaining two positions being filled in $2^{2} = 4$ ways. This gives us $4$ more words.

Thus there are $32 + 80 + 48 + 4 = \boxed{164}$ possible words.

Let the word be of n letters and their arrangements be represented as $a_{n}$

There might be two cases,

$\color{#D61F06}{\text{Case 1}}$

$\color{#3D99F6}{\text{Last letter is x}}$

Now the second last letter may be y or z . So the number of such rearrangements= $2a_{n-2}$

$\color{#D61F06}{\text{Case 2}}$

$\color{#3D99F6}{\text{Last letter is y or z}}$

Now there are no restrictions on second last letter. So the number of such rearrangements would be $2a_{n-1}$

$a_{n}=2a_{n-1}+2a_{n-2}$ $a_{5}=2a_{4}+2a_{3}$ $\Rightarrow a_{5}=6a_{3}+4a_{2}$ $\Rightarrow a_{5}=16a_{2}+12a_{1}$

Now $a_{2}$ can easily be calculated as $8$ and $a_{1}$ is obviously $3$ .

So $a_{5}=\boxed{164}$

Let $W_{n}$ denote the number of words allowed of length $n$ . Prove that it follows the recurrence relation(yeah, another 'Fibonnaci series' problem :P): $W_{n} = 2W_{n-1} + 2W_{n-2}$ From here you get the characteristic equation $x^2-2x-2=0$ with roots $1+\sqrt{3}$ and $1-\sqrt{3}$ which will give you the general solution of $W_{n} = a(1+\sqrt{3})^n + b(1-\sqrt{3})^n$ Try the easier one: Bash Me Not