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?
 A: There are ${8 \choose 2}$ ways to place the threes, then ${6 \choose 3}$ ways to place the ones, and then ${3 \choose 3}$ to place the remaining sixes.
All together we can arrange these groups in
$${8\choose 2}\cdot{6\choose3}=\frac{8!}{2!\cdot6!}\cdot\frac{6!}{3!\cdot3!}=\frac{8!}{2!\cdot3!\cdot3!}$$
different ways.
Notice that this is a multinomial distribution which takes the form
$$P(X_1=x_1,...,X_k=x_k)=\frac{n!}{x_1!\cdots x_k!}p_1^{x_1}\cdots p_k^{x_k}$$
Let $X_3,X_1,X_6$ denote the numbers of threes, ones, and sixes observed, respectively. Then
$$P(X_3=2, X_1=3,X_6=3)=\frac{8!}{2!\cdot3!\cdot3!}\left(\frac{1}{6}\right)^8\approx3.334\cdot10^{-4}$$
where we do not have to take into account $X_2=X_4=X_5=0$ since $0!=1$ and $\frac{1}{6}^0=1$
R Simulation:
dice=c(1,2,3,4,5,6)
u = replicate(10^7,sample(dice,8,repl=T))
one=colSums(u==1)
two=colSums(u==2)
three=colSums(u==3)
four=colSums(u==4)
five=colSums(u==5)
six=colSums(u==6)
mean(three==2 & one==3 & six==3)

[1] 0.0003365

which agrees with our result fairly accurately.
A: 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 
$$P = \frac{8!}{2!\cdot 3! \cdot 3!}\cdot \frac{1}{6^8}$$
