$15 \cdot 5$ There is a set containing three positive integer elements $\{a, b, c\}$, if $a \times b \times c=$ 2310, then the number of such sets is
(A) 32.
(B) 36.
(C) 40.
(D) 43.
(E) 45.
(46th American High School Mathematics Examination, 1995)
[Solution] Since $2310=2 \times 3 \times 5 \times 7 \times 11$, it has five prime factors.
5 factors and their products (partial products) can each be one of $a, b, c$, so the total number of sets is
$$
C_{5}^{2} C_{3}^{2}+C_{5}^{3}+C_{5}^{4}=45 \text{ (sets). }
$$

Noting that the elements $a, b, c$ should be distinct, the set $\{1,1,2310\}$ is clearly not included. Therefore, the answer is $(E)$.