Take the 2-minute tour ×  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: