Q:

combinatorics dice question

There are $10$ identical dice ($1$ - $6$).
How many different results can we get so that the set of results will be exactly $3$.
for example: $7$ dice will be the number $2$, $2$ dice will be $3$ and the last die will be $6$ so the set of results is $\{2,3,6\}$.
We are arguing here over the answer, I think that the answer is 
$$\frac{6 \cdot 5 \cdot 4 \cdot 3^7}{2}$$
while my friend strongly suggests that we will have to use the inclusion - exclusion principle. 
Thank you in advance

A:

if the order  does not matter:first select the three numbers you want to appear in $\binom{6}{3}$ ways. Once this has been done you want to divide the ten results into three groups. To do this use stars and bars, there are $10$ stars, and $2$ bars. But since every part needs at least one star there are really $7$ stars and so the answer is $\binom{9}{2}$
Hence you want $\binom{6}{2}\binom{9}{2}$.

If order matters: What you want to count is functions from $\{1,2,3,4,5,6,7,8,9,10\}$ to $\{1,2,3,4,5,6\}$ that have a range with three elements. There are $\binom{6}{3}$, once you have selected the range you want to count the number of surjective functions from $\{1,2,3,4,5,6,7,8,9,10\}$ to that range. to do this apply the twelvefold method. The answer is $3! {10 \brace 3}$. Hence the final answer is $\binom{6}{3}3!{10\brace3}$.
Note $\binom{n}{k}$ is the binomial coefficient and $n\brace k$ is the stirling coefficient of the second kind..

