Here's my solution, but as usual with combinatorics problems, I tend to be convinced of my errors too early. So I'd like to know what you guys think. Is this correct?
$\frac{\binom{10}{2} \binom{10}{3} 4^5}{6^{10}}$
Justification: There are $6^{10}$ ways of rolling of 10 dice. Now there are 10 choose 2 ways of having exactly two 2's among the dice, and 10 choose 3 ways of having three 3's. In addition, the remaining 5 dice may have any values except 2 or 3, which is $4^5$ ways of choosing them.
Right, right?
Best Answer
Almost.
Once you have selected 2 of the 10 places for the twos, there are only 8 places remaining; then select 3 of these for the threes.
$$\dfrac{\dbinom{10}{2}\;\dbinom{8}{3}\;4^5}{6^{10}}$$