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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.

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In (3), you mean $3 \cdot 3^2=27$. – Chris Eagle Jan 20 '12 at 21:51
There is also $1$ way the number $1$ appears three times face down, and the fact that $27+27+9+1=64$ should be encouraging. – Henry Jan 20 '12 at 21:55
That sounds right. As a check, you should argue that $1$ can appear on the bottom on no rolls, one roll, two rolls, or three rolls (there are no other possibilities), and you have worked out the number of ways that $1$ never appears, appears once, and appears twice for a total of $27 + 27 + 9 = 63$ of the $64$ occurrences. The remaining possibility is that of $1$ on all three rolls, right? Of course, all the numbers adding up doesn't guarantee that they are correct, but if they don't add up, that suggests that the work and the argument needs to be checked. – Dilip Sarwate Jan 20 '12 at 21:57
@Dilip You should make that an answer. – David Mitra Jan 20 '12 at 21:59
The work was nicely done. – André Nicolas Jan 20 '12 at 22:48

Yes it is correct, well done.

An alternate/complimentary way to answer question 4 is: If 2 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: C(N,1) = C(N, N-1) = N . In other words I can choose 1 from N, or complimentary (and exactly equivalently) I can choose N-1 from N. 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 confirms it.

And more general C(N, i) = C(N, N-i) You might already know that, but maybe now you have 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?

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