Let it be a tetrahedron with the numbers $1$,$2$,$3$ and $4$ on its faces.The tetrahedron is launch $3$ times. Each time, the number that stays face down is registered.
$1$)In total how many possible ways there are to registered the $3$ launches?
As there are $4$ numbers to $3$ launches(positions), the order matters and each number can repeats itself. I used a permutation with replacement: $4^3=64$
$2$)How many possible ways there are to the number $1$ never face down?
In this case I reduced the sample set to $\{2,3,4 \}$, and made the same as before. But this time there are $3$ numbers to $3$ launchs: $3^3=27$
$3$)How many possible ways there are to the number $1$ appears only $1$ time face down?
There are $3$ ways for number $1$ can be put on the $3$ launches. For the $2$ left there is $\{2,3,4 \}$.So I made a permutation with replacement: $3 \cdot 3^2=27$
$4$)How many possible ways there are to the number $1$ appears exactly $2$ times face down?
First I made a combination: $C(3,2)$ to find the number of ways that the pair of $1$'s can be put in the $3$ launchs.Then I multiplied by $3$, that is $ \{2,3,4 \}$ : $C(3,2) \cdot 3=9$.
Is this correct? Thank you very much, you(plural)have been very helpful.