Solution:
The answer is (B). Let $m$ be the number of Mathematicians and $n$ the total number of incoming students.
If for each freshman we consider the group consisting of that person, his right-hand neighbor, his left-hand neighbor and the person opposite him, we get $n$ groups, each of which contains exactly one Mathematician. On the other hand, each Mathematics freshman appears in exactly 4 of these groups (those corresponding to herself, to each of the two neighbors, and to the student seated opposite), from which $n=4 m$.

Let us now consider four people sitting next to one another counterclockwise, which we call $a_{1}, a_{2}, a_{3}, a_{4}$. Without loss of generality we can say that the first three are, in order, of groups $F, M$, and $C$. The person $a_{4}$, however, could be $F$ or $B$: let us now show that it is necessarily the second case.

Let us call $b_{i}$ the person sitting opposite $a_{i}$. Then $b_{2}$ is necessarily $B$. If we consider $b_{3}$, who sits next to $b_{2}$ counterclockwise, it cannot be $B$ or $M$, because of the conditions given by $b_{2}$, but neither can it be $C$, since it sits opposite $a_{3}$, which is $C$. Hence $b_{3}$ is $F$ and, similarly, $b_{1}$ is $C$. Let us now look at $b_{4}$: from the conditions on $b_{3}$ we must exclude $F, C$ and $B$, so it can only be $M$. But then $a_{4}$, who sits opposite it, cannot be $M$ and must be $B$.

So choosing the type of three people also fixes the fourth and, consequently, everything else. However, we must check that the patterns "fit together" well, since we must respect the condition on who sits opposite whom.

Let us number the seats around the table from 0 to $4 m-1$. The person sitting opposite 0 is $2 m$ and, in general, the person sitting opposite $k$ is $2 m+k$. We observe that people whose seat numbers lie in the same residue class modulo 4 have freshmen of the same type. Hence it is necessary that the residue class modulo 4 of $k$ and $2 m+k$ be different. If $m=2 h$, then $2 m+k=4 h+k \equiv_{4} k$, absurd. If instead $m=2 h+1$, then $2 m+k=4 h+2+k \equiv_{4} k+2$. The people at the table, therefore, can be in a number that is a multiple of 4 but not of 8. Between 30 and 50 the numbers with these characteristics are 36 and 44, whose sum gives 80.