10 dice are rolled. What is the probability of getting 6 dice that are even numbers and the other four dice are 3's? Ten dice are rolled. What is the probability of getting six dice that are even numbers and the other four dice are $3$'s?
My approach is that you have $6^{10}$ number of total outcomes, and 


*

*$\Pr(\text{6 dice with even numbers}) = (1/2)^6$

*$\Pr(\text{the other 4 with 3's}) = (1/6)^4$
So, the probability of the question will be $\frac{(1/2)^6 \cdot (1/6)^4}{6^{10}}$
But I am not confident that this is correct.

Thank you Arthur for correcting me, I got a bit carried away. It's night-time here where I'm at, after all.
Out of the 10 dice you roll, 4 need to be 3's (the other 6 will be even and uniquely determined, then), but there are many ways in which this condition can be satisfied. For example, you can roll $(3, 3, 3, 3, 2, 2, 2, 2, 2, 2)$, or you can roll $(3, 2, 3, 2, 3, 2, 3, 2, 2, 2).$ All in all, there will be $\binom{10}{4}$ such combinations.
The final solution, therefore, if we denote $A$ as the set of all outcomes where the conditions are satisfied, is $$P(A)={\binom{10}{4}(\frac{1}{2})^6(\frac{1}{6})^4}.$$
Edit 2: Again, thank you. I think it should be correct now, finally.
