A fair 6-sided die is rolled 6 times, what's the probability the outcome has exactly 2 or 3 elements? A fair $6$-sided die is rolled $6$ times independently. For any outcome, this is the set of numbers that showed up at least once in the different rolls. For example, the outcome is $(2,3,3,3,5,5)$, the element set is $\{2,3,5\}$. What is the probability the element set has exactly $2$ elements? how about $3$ elements? 
I know the sample space is $6^6$. The counting for $2$ elements is $6C2 \cdot 6!$? I would really appreciate the help! :)
 A: Exactly two numbers appear:
There are $\binom{6}{2}$ ways for two of the six numbers to appear.  On each of the six rolls of the die, there are two possible outcomes, giving $\binom{6}{2} \cdot 2^6$ possible outcomes in which two of the six numbers appear.  However, of these $2^6$ sequences, there are two in which the same number is rolled six times.  Thus, the probability of exactly two numbers appearing in six rolls of the die is 
$$\frac{\binom{6}{2} \cdot (2^6 - 2)}{6^6}$$
Exactly three numbers appear:
There are $\binom{6}{3}$ ways for three of the six numbers to appear.  On each of the six rolls of the die, there are three possible outcomes, giving $3^6$ sequences containing these three numbers.  However, we have counted the sequences in which not all three of the numbers appear.  There are $\binom{3}{2}$ ways for two of the numbers to appear.  For each such pair, there are $2^6$ sequences, of which two consist of sequences in which one number is rolled all six times.  Thus, there are $\binom{3}{2}(2^6 - 2)$ sequences in which exactly two of the three numbers appear.  The number of sequences in which exactly one of the three numbers appears is $3$.  Hence, the probability that exactly three of the numbers appear is 
$$\frac{\binom{6}{3}[3^6 - \binom{3}{2}(2^6 - 2) - 3]}{6^6}$$ 
My thanks to @bof for clarifying my thinking about the first problem.  Any errors that remain are entirely my responsibility.
A: For two groups:
There are $6\choose 2$ ways of selecting the 2 groups. There are ${6\choose 2}-2$ ways to for 6 dice to roll those numbers.
Can you work it out for 3 groups?
A: Why can't we use the multinomial formula? We want to have a sequence of 2 repeated elements 3 each. This sounds like $\frac{6!}{3!*3!}$, right? 
For example we have {1,1,1,2,2,2} or a combination of this. Then we get $\frac{6!}{3!*3!}$ since we have 2 sets of 3 repeated elements and then we divide by the sample space times 6 (there are six possibilities). The result seems to be incorrect using this method. I tried using a 3^3 tree and it seemed ok. Any thoughts?
