We throw a die $8$ times. What is the probability of obtaining exactly two $3$s, three $1$s, three $6$s? We throw a die $8$ times.  What is the probability of obtaining exactly two $3$s, three $1$s, three $6$s? 
My work:
The sample space  $ S:$  "The set of all solutions of throwing the die [eight times]" and $|S|:6^8$
Then the probability of getting two $3$s, three $1$s, three $6$s is $\frac{8}{6^8}$
Is the reasoning correct?

To find the number of ways to get two 3s, three 1s and three 6s can also be seen as a simple permutation problem of $8$ digits.
Imagine the result of the $8$ throws is recorded as a sequence of $8$ numbers: $d_1 d_2 \ldots d_8$.
Now you want to know in how many different ways can occur the numbers $3,3,1,1,1,6,6,6$ in such a sequence. 
Or, in other words, how many different 8-digit numbers can you create with the digits $3,3,1,1,1,6,6,6$.


*

*there are $8!$ permutations (arrangements) of these $8$ digits but

*each permutation of $3,3$ and $1,1,1$ and $6,6,6$ gives the same 8-digit number, so you get


$$\frac{8!}{2!\cdot 3! \cdot 3!} \mbox{ differnt 8-digit numbers or different throws}$$
So, the probability in question is 
$$P = \frac{8!}{2!\cdot 3! \cdot 3!}\cdot \frac{1}{6^8}$$
