Probability question with similar/distinct dice. The question: 

We're throwing $3$ normal dice twice. (Suppose that every vector with length of $6$ has an equal probability). What is the probability to get two similar results from the two throws of the $3$ dice if: 
a) The dice are different. 
b) The dice are similar.

My Work: 
for (a):  $\Omega = \{(a_1,a_2,...,a_6) \mid a_i \in \{1,...,6\}  ,(i=1,2,...,6)\}$
if all the dice are different then: $|\Omega| = 6^6$ 
Let $A\subset\Omega$, ($A$- we get similar results from the two throws), then $|A|=6^3$ (first dice throw can be whatever, but the second one must be the same). 
So the final answer I got is $P(A) = \frac {6^3}{6^6}$. 

for (b): (and here comes the confusion): 
Things I tried to do and didn't get the right answer: 

*

*I tried to choose $3$ of the dice (since they're all similar) throw them, and sort them, so what I got is $\binom{6}{3} \cdot 6^3 \cdot 3!$. 

*I tried to choose $1$ of the two throws, and find how many options there are and sort them (same as before, each dice has $6$ sides so $6^3$ for all), and got $\binom{2}{1} \cdot 6^3 \cdot 3!$. 

Both of the two ways led me to wrong answers, I'm trying where are my mistakes and why my thinking led me to wrong answers, which can lead me to reach the true answer. 
The Final answer is $\frac {996}{6^6}$ 
EDIT: 
After deeply thinking about the problem I obtained a new approach (I couldn't reach the answer with it, but I'm not sure if it's my weak combinatorics or if it's a wrong approach). 

*By writing down a couple of examples, I noticed that there's something that I can also use to calculate the probability, we can divide this into $3$ possible cases: 
First: Getting same number on all of the six dice. $(2,2,2, 2,2,2)$
Second: Getting $2$ different numbers. $(2,4,2, 2,4,2)$ 
Third: Getting $3$ different numbers. $(1,2,3, 1,2,3)$ 
So I could calculate the number of options each case has and divide by the total options $|\Omega|=6^6$.

I felt like I've gone too far but I can't see why it wouldn't be true. Any Feedback is really appreciated.

 A: With the help of N. F. Taussig in the comments, I think I have reached the right solution, So I decided to answer my own question, and would love to hear feedback. 

Starting with why my first two approaches are wrong, The number of ways that the second roll could match the first is different whenever there's $3$ different numbers or $2$ different or all of them are the same, but in my first two approaches, I didn't pay attention to that. 

Solution: (Approach 3): 
The number of ways to get the same number on all dice, is just simply choosing a number which is $\binom{6}{1}=6$. 
The number of ways to get two different numbers in the first roll is:  $\binom{6}{2}$ choosing $2$ numbers, $\binom{2}{1}$ choosing the number that will appear twice, $\binom{3}{1}$ choosing which place is the single number in the first roll, $\binom{3}{1}$ choosing in which place is the single number in the second roll. So in total by multiplying all of them we get $\binom{6}{2}*2*3*3 = 270$ ways. 

The number of ways to get three different numbers in the first roll is: 
$\binom{6}{3}$ choosing $3$ numbers, $3!$ ordering the first roll, $3!$ ordering the second roll.  in total $720$. 

Hence we reach the answer by adding them up and dividing by $|\Omega|$.
