 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. HenceIn the order $(X_1,\dots ,X_9)$, the two forms arewhere $(a,b,c)$ ranges over the six permutations of $(0,1,2)$.