Probability and combinatorics on tetahedron faces. 
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.
 A: Yes it is correct, well done.
An alternate/complimentary way to answer the last question:
If two throws are '1' then we have only one throw which is not '1'. There are 3 ways for the not-'1' throw to be put down. And not-'1' has 3 cases as you said: $\{2,3,4\}$. So again $3 \cdot 3 = 9$
This might make you notice that in general: $_NC_1 = \space _NC_{N-1} = N$. In other words I can choose $1$ item from $N$ items in $N$ different ways, or complimentary (and exactly equivalently) I can choose $N-1$ items from $N$ items. So if we are throwing the tetrahedron $N$ times you could either ask: 1) how many ways to roll '1' (N-1) times?, or 2) how many ways to roll not-'1' one time? It's the same question isn't it? And combinatorics rightly produces the same answer to both of them.
More generally $_NC_i = \space _NC_{N-i}$. You might already know that, but maybe now you can get some more insight into it.
Can you give the general solution to you problem? That is, let's say we have a set of $N$ numbers (a dice of $N$ sides) and we throw it $M$ times. How many possible ways exist to get the same number $K$ times?
