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.
So you have a total of $26$ people to choose from, teachers and students but only $5$ can be representatives for the team. We need to keep in mind that we can't have all teachers or students in the committee at one time. So what this means is that the number of ways to arrange the teachers is $$C (12,5) = 792$$ The number of ways to choose the students is $$C(14,5) = 2002$$ This is straight from what we are given. Overall, if we assume that we have no restriction at all, we have a total of $$C(26,5) = 65780$$ Since we can't have all teachers or students represent the team, subtract $$65780 - 2002 - 792 = 62986$$ Another reason we subtract from the total is because we don't want to count our arrangements more than once.
Best Answer
You can do so by dividing them into cases:
Case I: Three teachers and one student
You can choose this in $\binom{8}{3} \times \binom {7}{1}$ ways.
So the probability for case I is given by $\frac{56 \times 7}{1365}=\frac{392}{1365}$
Case II: Four teachers and no student
You can choose this in $\binom{8}{4}$ ways.
So the probability for case II is $\frac{70}{1365}$
The final probability will be found by adding the two cases.