Think of it this way. Arrange the students in a line (there are $17!$ ways to create this permutation):
$$
\begin{array}{ccccccccccccccccc}
s_1 & s_2 & s_3 & s_4 & s_5 & s_6 & s_7 & s_8 & s_9 & s_{10} & s_{11} & s_{12} & s_{13} & s_{14} & s_{15} & s_{16} & s_{17} \\
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17
\end{array}
$$
Then create the groups by taking the first 4, second 4, third 4 and last 5:
$$
\begin{array}{cccc|cccc|cccc|ccccc}
s_1 & s_2 & s_3 & s_4 & s_5 & s_6 & s_7 & s_8 & s_9 & s_{10} & s_{11} & s_{12} & s_{13} & s_{14} & s_{15} & s_{16} & s_{17} \\
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17
\end{array}
$$
Every way to group the students can be created this way. The key question is: how many permutations lead to the same set of groups?
N. B. we want the bars in these exact positions. If we allow ourselves to place the bars differently we would have to multiply $17!$ by $4$ to include the extra data of where the group of $5$ is taken from (first, second, third or fourth group).
We see of course that we can permute the people in any of the 4 groups and it doesn't change the group. Example:
$$
\begin{array}{cccc|cccc|cccc|ccccc}
s_3 & s_2 & s_4 & s_1 & s_6 & s_5 & s_7 & s_8 & s_9 & s_{10} & s_{12} & s_{11} & s_{14} & s_{13} & s_{17} & s_{15} & s_{16} \\
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17
\end{array}
$$
There are $4!$ ways to permute each of the first three groups and $5!$ ways to permute the fourth group. This brings us down to $\frac{17!}{4!4!4!5!}$.
Also note, that when we do this, there is an inherit order to the three groups of $4$. Specifically, things like this:
$$
\begin{array}{cccc|cccc|cccc|ccccc}
s_5 & s_6 & s_7 & s_8 & s_1 & s_2 & s_3 & s_4 & s_9 & s_{10} & s_{11} & s_{12} & s_{13} & s_{14} & s_{15} & s_{16} & s_{17} \\
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17
\end{array}
$$
is considered different. There are $3!$ ways to permute the groups of $4$ leading to the final answer of
$$ \frac{17!}{4!4!4!5!}/3!. $$
Note that we cannot swap a group of $4$ with a group of $5$ because that would require moving the bars. The goal we started with is to count how many of the $17!$ permutations lead to the same groups of students by placing the bars after the fourth, eighth and twelfth student.
Yes, your explanation is absolutely correct. However, I would like to present even a simpler explanation for this problem.
Imagine having a dinner table with 60 seats. The table is divided into 2 sections of 20 seats and 40 seats. See the diagram below
Your problem essentially is to place a particular candidate to 20-seat-section. And you don't care how the rest of other candidates are seated.
Probability of the person getting seated to 20-seat-section is simply $\frac{20}{60}$ = $\frac{1}{3}$
Best Answer
Imagine that we line up the students at random, and assign the first $4$ to one group, the next $4$ to another, and the last $4$ to another.
We can analyze as follows. We have $3$ copies of the letter $H$, and $9$ copies of the letter $M$. We make a $12$-letter word. What is the probability there will be an $H$ among the first $4$ letters, and an $H$ among the next $4$, and an $H$ among the last $4$?
There are $\binom{12}{3}$ equally likely ways to place the $3$ $H$'s.
The number of choices with one $H$ in the first $4$, another $H$ in the next $4$, another in the next $4$ is $4^3$.
Our required probability is therefore $\dfrac{4^3}{\binom{12}{3}}$.
Remark: Your number $\binom{12}{8}\binom{8}{4}\binom{4}{4}$ counts the number of ways to divide our $12$ people into three uniformed teams, the Blues, the Whites, and the Reds. That's fine.
Now we count the number of ways to divide into uniformed teams with an Honours student on each team. The Honours student of the Blues can be chosen in $\binom{3}{1}$ ways. The rest of the team can be chosen in $\binom{9}{3}$ ways. For every such choice, the Honours student in the Whites can be chosen in $\binom{2}{1}$ ways, and the rest of her team in $\binom{6}{3}$ ways. Now we are finished, although we can tack on a decorative $\binom{1}{1}\binom{3}{3}$ for a total of $\binom{3}{1}\binom{9}{3}\binom{2}{1}\binom{6}{3}\binom{1}{1}\binom{3}{3}$.
Now for the probability, divide.