Faces of a dice

Total ways to choose $3$ alphabets $\{x, y, z\}$ from $\{A, B, C, D, E, F\}$ $=$ $6 \choose 3$

For a given triplet $\{x, y, z\}$ ,

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Total ways to arrange alphabets from $\{x, y, z\}$ in the list $= 3^{15}$

$$

Total ways to arrange $2$ alphabets from $\{x, y, z\}$ in the list $= 3 \times 2^{15}$

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Total ways to arrange $1$ alphabet from $\{x, y, z\}$ in the list $= 3 \times 1^{15}$

$$

Hence, total ways to arrange alphabets from $\{x, y, z\}$ in the list

such that each of the alphabet appears at least once $= 3^{15} - 3 \times 2^{15} + 3 \times 1^{15}$

$$

Hence, total required ways $=$ $6 \choose 3$ $\times (3^{15} - 3 \times 2^{15} + 3 \times 1^{15}) = 285012120$