A fair die is rolled eight times. What is the probability of getting exactly 2 threes, exactly 3 ones, and exactly 2 sixes? I was given the following hint: 
Break the task of counting the desirable outcomes into three subtasks: 


*

*(a) choose the location of the extra non-three, one, or six in the sequence of eight throws,

*(b) permute the seven threes, ones, and sixes, and

*(c) choose a non-three, one, or six to occupy the location determined in subtask (a).


However, I am confused by this hint. From my understanding, in order to combine these three tasks, I would do:
$$8 \times (2!) \times  (2!) \times  (2!) \times \frac{\left(\frac 12\right)^8}{8!}$$
because there are 8 possible locations for the extra non-three, one, or six; the permutations of the threes, ones, and sixes are $2!, 2!$, and $3!$; and the probability of not rolling a one, three, or six 8 times is $\left(\frac 12\right)^8$. But this doesn't seem correct for some reason. Disregarding the hint, I would approach the problem by doing the following:
$$8 \times \left(\frac 16\right)^7 x \left(\frac 56\right)$$
I can't really tell which approach is correct, because I'm having a really hard time visualizing this problem.
 A: Record the outcome of the rolling as a string of length $8$, made  of symbols chosen from $\{1,2,3,4,5,6\}$. There are $6^8$ such strings, all equally likely if the die is fair and rolled fairly.
Now we count the number of good strings, where good means fitting the specifications in the title.
The location of the digit which is different from $3$, $1$, or $6$ can be chosen in $8$ ways. For each choice of location, there are $3$ ways to choose which one of the three possibilities the  "odd" digit will be. Thus there are $(8)(3)$ ways to decide on the location and type of the "odd" digit.
For every way of doing the above two tasks, there are $7$ empty slots. There are $\frac{7!}{2!3!2!}$ ways of filling them with two $3$'s, three $1$'s, and two $6$'s. That gives a total of $(8)(3)\left(\frac{7!}{2!3!2!}\right)$ good strings. Divide by $6^8$ for the probability.
We can count the number of ways of filling the $7$ slots more slowly. The location of the $3$'s can be chosen in $\binom{7}{2}$ ways. For each of thse choices, the location of the $1$'s can be chosen in $\binom{5}{3}$ ways, for a total of $\binom{7}{2}\binom{5}{3}$. That gives the expression $(8)(3)\binom{7}{2}\binom{5}{3}$ for the number of good strings.
Remark: There are other ways to organize the count. For example, let us make an $8$-letter string, made up of two $3$'s, three $1$'s, two $6$'s, and an $x$, to act as a placeholder for the "odd" number. There are $\frac{8!}{2!3!2!}$ ways to do this. Multiply by $3$, because $x$ can be replaced by any of $3$ values.
A: Let $D_k$ be the event that the die turns up $k$. Let $D_{i_1}, D_{i_2}, \cdots D_{i_k}$ denote the event that the first roll turns up $i_1$,the second $i_2$, etc.
First compute $D_3 D_3 D_1 D_1 D_1 D_6 D_6 D_k$, where $k \in \{ 2, 4, 5\}$. The probability of this specific event for a particular $k$ is $\frac{1}{6^8}$, and the combined probability of the (mutually exclusive) events for $k \in \{ 2, 4, 5\}$ is given by $3 \frac{1}{6^8}$.
Now see how we can rearrange the events. First select the positions for the 3's, there are $\binom{8}{2}$ possibilities. Now select the 1's, there are $\binom{6}{3}$ possibilities, and finally select the 6's, there are $\binom{3}{2}$ possibilities. Note that $\binom{8}{2} \binom{6}{3} \binom{3}{2} = \frac{8!}{2!\, 3!\, 2!}$.
Hence the answer is $3 \frac{1}{6^8} \frac{8!}{2!\, 3!\, 2!} = \frac{35}{11664}$.
