= Solution
Choose the distinct values of $(X_1,X_2,X_4)$ and call them $(a,b,c)$. There are $3!=6$ choices. Constraint (5), on $(X_1,X_4,X_9)$, immediately forces $X_9=b$.
Constraints (2) and (3) require $X_5\ne X_2=b$ and $X_5\ne X_4=c$, respectively. Since the <constraint domain> has exactly three values, $X_5=a$. Constraint (2) now forces $X_3=c$, and constraint (3) forces $X_6=b$. Finally, constraint (4) permits $(X_7,X_8)=(a,c)$ or $(c,a)$.
Every choice just described satisfies all five <all-different constraints>, and the forced-value argument shows there are no other possibilities. Hence
$$
\boxed{\#\operatorname{Sol}(P)=3!\cdot2=12.}
$$
In the order $(X_1,\ldots,X_9)$, the two forms are
$$
(a,b,c,c,a,b,a,c,b),\qquad
(a,b,c,c,a,b,c,a,b),
$$
where $(a,b,c)$ ranges over the six <permutations> of $(0,1,2)$.
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