Probabillity - rolling a dice 20 times, probability of a result gets only once 
A dice is rolled $20$ times, with the possible results $\left\{1,2,3,4,5,6\right\}$.
Let $X$ be the number of results, out of the possible 6, which were chosen only once during the 20 rolls.
Calculate $P\left\{X\right\}$

I find it hard to identify the kind of variable it is. It isn't bio nominal nor hyper geometric.
I understand I have to choose 4 rolls out of the 20, and the combination between them is $4!$, giving me -
$$ \frac{\binom{20}{4} \times \binom{6}{4} \times 4!}{6^{20}} $$
For the chosen "results", the chosen "rolls' and the inner combination between them. But how about the other "rolls"? Something is missing here.
 A: Observe that since there are $20$ rolls, $P(X=6)=0$. So we need only check the probabilities that $X=1,2,3,4,5$. These can be done on a case by case basis.
For $X=5$, the probability is
$$\frac{\binom{6}{5}\cdot\binom{20}{5}\cdot 5!}{6^{20}}$$ 
since we must choose the the $5$ results which will occur only once, and choose which rolls they occur on in $\binom{20}{5}$ ways, accounting for their orderings.
For $X=4$, the probability is
$$\frac{\binom{6}{4}\cdot\binom{20}{4}\cdot 4!\cdot (2^{16}-30)}{6^{20}}$$
since we must choose the $4$ results which will occur only once, choose which rolls they occur on, order these $4$ results (in $4!$ ways), and then fill in the remaining $16$ rolls with the other two results. We must be a bit careful here, since we need each of the other results to occur at least twice, or not at all. Denote the remaining results by $x$ and $y$. Since there are $16$ rolls to fill in with $x'$s and $y$'s, there are $2^{16}$ possible outcomes. $15$ of them consist of $15 x'$s and one $y$, and another $15$ consist of $15 y$'s and one $x$. Discarding these $30$ undesirable outcomes leaves $2^{16}-30$.  
The argument is similar for the other cases, but the last bit corresponding to the results that don't appear exactly once gets a bit more complicated. For $X=3$, we have have $17$ rolls that must be filled with, say, $x,y,z$ such that neither $x,y,$ nor $z$ appears exactly once. There are $17\cdot 2^{16}$ ways for $x,y,$ or $z$ to appear once (place it in one of $17$ positions then fill the other $16$ rolls with the other two results). And there are $\binom{17}{2}$ ways for two of them to appear only once. Since these are double-counted above, there are $3(17\cdot 2^{16}-\binom{17}{2})$ undesirable cases to discard. 
I'll leave the cases $X=1,2$ up to you to compute. For now I'll just denote by $C_{4},C_{5}$ the number of ways to arrange the remaining results without any of them appearing exactly once. Therefore
$$P(X=3)=\frac{\binom{6}{3}\cdot\binom{20}{3}\cdot 3!\cdot [3(17\cdot 2^{16}-\binom{17}{2})]}{6^{20}}$$ 
$$P(X=2)=\frac{\binom{6}{2}\cdot\binom{20}{2}\cdot 2!\cdot (4^{18}-C_{4})}{6^{20}}$$
$$P(X=1)=\frac{\binom{6}{1}\cdot\binom{20}{1}\cdot (5^{16}-C_{5})}{6^{20}}$$ 
For completeness, observe that (clearly) $P(X<1)=P(X>6)=0$.
