Answer
$\color{blue}{p_X(k) = \begin{cases} 1/36,& k =1 \\ 15/36, & k=2 \\ 20/36, & k=3\end{cases}}$
Work Step by Step
The total number of outcomes for three (3) tosses of a fair dice is $6^3=216$ as each toss may come up in $6$ ways (the face shown for each toss is one of $1,2,3,4,5$, or $6$).
If we let the random variable $X$ denote the number of different faces that appear in three tosses of a fair die, then $X$ can either be $1$ (if all three dice show the same face), $2$ (if exactly two dice show the same face), or $3$ (if all three dice show different faces).
The number of ways in which $X=1$ (i.e., all three tosses of the fair die show the same face) is $6$ ways, since there are $\displaystyle {6\choose 1} = 6$ ways to choose the number that will show on each of the three tosses of the fair die.
Thus, $P(X=1) = 6/216=1/36$.
The number of ways in which $X=2$ (i.e., exactly two of the three tosses show the same face) is $\displaystyle {6\choose 1}\cdot {3\choose 2}\cdot 5 = 6\cdot 3\cdot 5 = 90$ ways since there are $\displaystyle {6\choose 1} = 6$ ways to choose the number that will be shown by exactly two of the three tosses, there are $\displaystyle {3\choose 2} = 3$ ways to choose which two tosses will share this number, and there are $5$ ways to assign a number for the face shown on the remaining toss of the three tosses.
Thus, $P(X=2) = 90/216=15/36$.
The number of ways in which $X=3$ (i.e., all three tosses show different faces) is $6\cdot 5\cdot 4 = 120$ ways, since the first toss may come up in $6$ ways, the second toss in $5$ ways (as the face shown has to be different from that of the first toss), and the third toss in $4$ ways (as the face shown has to be different from that of the first two tosses).
Thus, $P(X=3) = 120/216=20/36$.
All told, the pdf of $X$ is
$\begin{align*}
p_X(k) &= P(X=k),\ k=1,2,3 \\
\color{blue}{p_X(k)} &\color{blue}{=\begin{cases} 1/36, & k=1 \\ 15/36, & k=2 \\ 20/36, & k=3 \end{cases} }
\end{align*}$
