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
On
Your calculation assumes that the first three dice have different values. The factor $6$ says the first can be anything, the $5$ says the second has to be different from the first, the $4$ says the third has to be different from the first two. The $3^7$ says each other die can be any of the three numbers. I don't understand the division by $2$. Following this logic you should divide by $6$ because the first three numbers can come in any order. You have missed rolls like $1111223333$ because the first three dice are the same.
I would use inclusion/exclusion. You can select the three numbers in ${6 \choose 3}=20$ ways. Then we want to find the number of $10$ digit strings from $1,2,3$ that include at least one of each number. There are $3^{10}$ strings total, then we subtract the $3 \cdot 2^{10}$ that have only two different digits, then add the $3$ that have a single digit. The final answer is ${6 \choose 3}(3^{10}-3\cdot 2^{10}+3)=1,119,600$
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..