How many ways can $26$ students be distributed. Consider the classrooms $a,b,c,d$ and suppose we have $26$ students. The first question is, how many ways can we distribute the $26$ students, if some rooms may be left empty? My answer to this question is $${4+26-1} \choose {26}$$.
Second question: In how many ways can we distribute the students if each classroom must contain at least $2$ students and at most $12$ students. My answer to this question is to find the coefficient to $x^{26}$ in the generating function $$(x^2+\ldots+x^{12})^4$$
Third question: How many ways can we distribute $9$ students to room $a$, $12$ students to room $b$, $15$ students to room $c$ and $3$ students to room $d$. My answer is $$ 26 \choose {9,12,15,3}$$
Are my answers correct, I feel I might have done something wrong.

In the first question you’ve treated the students as if they were indistinguishable, which is presumably not the case. Since we can tell one student from another, we care not only how many students end up on each room, but which students end up on each room. Thus, in the first problem we have a $4$-way choice of classrooms for each student, and we can therefore distribute the students amongst the classrooms in $4^{26}$ different ways.
Your second answer has the same problem as your first: there are $\binom{26}{6,6,7,7}$ different ways to assign $6$ students each to rooms $a$ and $b$ and $7$ each to rooms $c$ and $d$, but you’re counting all of those as a single distribution.
The third question is perhaps a bit of a trick question: $9+12+15+3>26$, so it’s impossible to distribute $9$ to room $a$, $12$ to room $b$, $15$ to room $c$ and $3$ to room $d$, and you can immediately write down $0$ for the answer. What you did write isn’t wrong, since that multinomial coefficient evaluates to $0$, but it’s less informative than it might be.
