NP Complete Clique Variation I have the following problem I am trying to prove NP Complete.

Tutors have two jobs- grading exams and homework help. Suppose we have a set of $n$ tutors. Each of them can tutor any amount of $m$ classes.

The tutors rent a large study room to offer help to students. But, note that ALL tutors for a specific class can't be in the study room. Otherwise, that class will have no one left to grade the exams.

Question is, can we have $k$ tutors in the study room?

Attempt at solution:

I feel like this is a set or vertex cover problem and have interpreted as follows.
We create a bipartite graph... connecting the left hand side (class) to right hand side (tutor). We have to see if we can choose $k$ tutors from the right hand side such that no left hand side vertex has all its edges used up.



For example, if $k = 1$ the answer is yes. Choose any person.
If $k = 2$ answer is yes. $\{A, D\}$ works but not $\{A, B\}$ because they make up the tutors for $C2$.
$k = 3, 4$ also do not work for the same reasoning.

I am trying to decide if I should approach this with set-covers, clique, independent set? I'm not entirely sure but feel this approach is on the right track.
You can prove this by a reduction from the Hitting Set problem (which is NP-Complete):


  Given a base set $U$, a family $F = \{S_1,\ldots,S_\ell\}$ of subsets of $U$ and an integer $q$, decide if there exists a set $H$ of size at most $q$ which intersects every set in $F$ (this is called a hitting set).


So suppose given an instance of the Hitting Set problem. For every element in $u\in U$, create a new TA $T_u$. For every set $S\in F$, create a new class $C_S$ whose TAs correspond to the elements in $S$. Let $k = |U|-q$. 

Then if you interpret the TAs being left out of the room as the elements chosen in $H$ it is not hard to see that there exists a hitting set of size at most $q$ if and only if you can put $k$ TAs in the room. That is because putting $k = |U|-q$ TAs in the room is equivalent to leaving at most $q$ outside of the room. And hitting every set in $F$ is equivalent to having at least one TA per class left out of the room.

(I leave the formal proof up to you)