Throwing a die three times - check my proof "A teacher rolls a 6-sided die (numbers from 1 to 6) three times. The lowest number will be the grade of the student. 
Calculate the probability of each grade."
My approach is:
Firstly we determine the sample space $\Omega$. So $\Omega:= \{\{\omega_1, \omega_2, \omega_3\},~with~\omega_i \in \{1,...,6\}\}$. The 3-sets $\{\omega_1, \omega_2, \omega_3\}$ are non-ordered and as repitition is allowed we have $\vert\Omega\vert={6+3-1\choose 3}$. 
1.) Let be $A_1$ the set of all 3-sets which at least contain one $1$. We have to consider the cases in which the 3-set contains one $1$, two $1's$ and three $1's$. Hence, $\vert A_1 \vert= {5+2-1 \choose 2}+5+1$. The probabilty of $A_1$ is $\frac{21}{56}= 0.375$.
2.) Let be $A_2$ the set of all 3-sets which at least contain one $2$ and no $1$. We have to consider the cases in which the 3-set contains one $2$, two $2's$ and three $2's$. Hence, $\vert A_2 \vert= {4+2-1 \choose 2}+4+1$. The probabilty of $A_2$ is $\frac{15}{56} \approx 0.268$.
In this manner I would proceed for the remaining 3-sets.
However, the solution comes up with slightly different probabilities as it uses the combinatorial model of "ordered sample" and "repitition allowed"
What is wrong with my approach?

The grade $1$ occurs when $1,2$ or $3$ dice show $1$. 
Only one die showing $1$:
$${3\choose 1}\cdot 5^2$$
Two dice showing $1$ each:
$${3\choose 2}\cdot 5$$
All three dice showing $1$:
$${3\choose 3}$$
Hence:
$$P(1)=\frac{{3\choose 1}5^2+{3\choose 2}5+{3\choose 3}}{6^3}\approx 0.421$$
Similarly:
$$P(2)=\frac{{3\choose 1}4^2+{3\choose 2}4+{3\choose 3}}{6^3}\approx 0.282\\
P(3)=\frac{{3\choose 1}3^2+{3\choose 2}3+{3\choose 3}}{6^3}\approx 0.171\\
P(4)=\frac{{3\choose 1}2^2+{3\choose 2}2+{3\choose 3}}{6^3}\approx 0.088\\
P(5)=\frac{{3\choose 1}1^2+{3\choose 2}1+{3\choose 3}}{6^3}\approx 0.032\\
P(6)=\frac{{3\choose 1}0^2+{3\choose 2}0+{3\choose 3}}{6^3}\approx 0.005\\$$
