For how many three-element sets of distinct positive integers $\{a,b,c\}$ is it true that $a \times b \times c = 2310$?

$\mathrm{(A) \ 32 } \qquad \mathrm{(B) \ 36 } \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 }$
$2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$. We wish to figure out the number of ways to distribute these prime factors amongst 3 different integers, without over counting triples which are simply permutations of one another.
We can account for permutations by assuming WLOG that $a$ contains the prime factor 2. Thus, there are $3^4$ ways to position the other 4 prime numbers. Note that, with the exception of when all of the prime factors belong to $a$, we have over counted each case twice, as for when we put certain prime factors into $b$ and the rest into $c$, we count the exact same case when we put those prime factors which were in $b$ into $c$.
Thus, our total number of cases is $\frac{3^4 - 1}{2} = 40 \Rightarrow \boxed{C}.$