Mass function 3 dice, Variance and Expected value. Experiment: 3 fair six dice are tossed.
X: # of distinct faces.
a)  Write the probability mass function of X.
b)  Calculate the expected Value and Variance of X
a)  In a fair dice I have (1 to 6, 1to 6, 1 to 6) possible outcomes. Then total outcomes are 6^3.
I notice 3 different cases,  1) 3 dice equal face  2) 2 dice equal face and 3) no face have equal values.
Therefore, 
1)  6 outcomes for 3 dice with face equal.   ->    X=0    P=6/6^3
2)  (1 to 6, 1 to 6, (5CN1)) for each dice then I have 18  outcomes for 2 dice with same face ->   X=2  P=18/6^3
3)  and (6^3 )- 36-6+1 total # of outcomes for the other possibilities.--> X=3 P=((6^3)-24)/6^3 
My question is regarding, how to make or construct the probability mass function with the data? And if my data and thought is corrected or not?

b)  I believe after I have the mass function, I will calculate the expected value and the variance with.
Expected value= X * fx(x)
              Will be:  E(x) 2.75
EV: {(6^3)-24)/6^3 }*3+{2 *(18/6^3)}+0 * {6/6^3}
Variance= the sum of all ((X-E(X))^2 )* P(X) =0.72
Is that correct? I proceed well? If not where is my mistake?
 A: The size of the total space is $6^3$ distinct (and equally probable) outcomes.
Case 1) All faces the same: $X=1$.   To obtain the size of the favoured space, count the ways to select one number to be repeated three times.   As you did. $$\mathsf P(X{=}1)= \frac{6}{6^3} = \frac{1}{36}$$
Case 2) Two faces the same, one different: $X=2$.   To obtain the size of the favoured space, count the ways to select a face for the pair, the ways to select a different face for the singleton, and the ways to choose which dice is a singleton.   You weren't quite right.
$$\mathsf P(X{=}2)=\frac{6\times 5\times {^{3}{\rm C}_{1}}}{6^3} = \frac {5}{12}$$
Case 3) All faces distinct: $X=3$.   Here we need to count the ways to select and arrange three distinct faces.
$$\mathsf P(X{=}2)=\frac{{^{6}{\rm C}_{3}}\times 3!}{6^3} = \frac {5}{9}$$
( Notice that the checksum is $1$.   $\frac{1}{36}+\frac{5}{12}+\frac{5}{9}=1$ )
These are the probability mass function.   $f_X(x)=\mathsf P(X{=}x)$
Now use these probabilities in the method you had.
$$\begin{align}\mathsf E(X) & = \sum_{x=1}^ 3 x\;\mathsf P(X{=}x)\\\mathsf {Var}(X) & = \sum_{x=1}^3 (x-\mathsf E(X))^2\mathsf P(X{=}x)\end{align}$$
alt: $\mathsf E(X^2) = \sum_{x=1}^3 x^2\,\mathsf P(X{=}x)$ and $\mathsf {Var}(X)=\mathsf E(X)^2 - \mathsf E^2(X)$
