Sample space be $S = \{(d_1, d_2, d_3) \ | \ 1 \le d_1,d_2,d_3 \le 6\}$ where $d_i$ correspond to $i^{th}$ dice.
$\underline{\textbf{Case 1:}} \text{ }$
$\forall_{1 \le j,k,z \le 3} \ j\neq k \land d_j=1 \land d_k=5 \land z \neq j \land z \neq k\Longrightarrow \forall_{2\le m\le 4} \ d_z = m$
In other way, we write it as $\underbrace{^3C_2}_{\text{selecting j, k}} \times \underbrace{2!}_{\text{arranging j,k}} \times \underbrace{^3C_1}_{\text{choosing }d_z} = 18$
And,
$\forall_{1 \le j,k,z \le 3} \ j\neq k \land d_j=2 \land d_k=6 \land z \neq j \land z \neq k\Longrightarrow \forall_{3\le m\le 5} \ d_z = m$
In other way, we write it as $\underbrace{^3C_2}_{\text{selecting j, k}} \times \underbrace{2!}_{\text{arranging j,k}} \times \underbrace{^3C_1}_{\text{choosing }d_z} = 18$
$\underline{\textbf{Case 2:}}$
$\forall_{1 \le j,k,z \le 3} \ j\neq k \land d_j=1 \land d_k=5 \land z \neq j \land z \neq k\Longrightarrow \forall_{m \in \{1,5\}} \ d_z = m$
In other way, we write it as $\dfrac{\underbrace{^3C_2}_{\text{selecting j, k}} \times \underbrace{2!}_{\text{arranging j,k}} \times \underbrace{^2C_1}_{\text{choosing }d_z}}{\underbrace{2!}_{\text{overcounting; e.g. (1,1,5) counted twice}}} = 6$
And,
$\forall_{1 \le j,k,z \le 3} \ j\neq k \land d_j=2 \land d_k=6 \land z \neq j \land z \neq k\Longrightarrow \forall_{m \in \{2,6\}} \ d_z = m$
In other way, we write it as $\dfrac{\underbrace{^3C_2}_{\text{selecting j, k}} \times \underbrace{2!}_{\text{arranging j,k}} \times \underbrace{^2C_1}_{\text{choosing }d_z}}{\underbrace{2!}_{\text{overcounting; e.g. (2,2,6) counted twice}}} = 6$
$\underline{\textbf{Total}}$
$\text{Favourable Cases} = 18+18+6+6 = 48$
$|S| = 6^3 = 216$
$\text{Probability} = \dfrac{48}{216} = \dfrac{2}{9}=0.\overline{2}$
