15. Among the divisors of 12!, if a positive integer is randomly selected, the probability that this divisor is a perfect square is $\frac{m}{n}$ (where $m$ and $n$ are coprime positive integers). Then the value of $m+n$ is ( ).
(A) 3
(B) 5
(C) 12
(D) 18
(E) 23
15. E.

Notice that, $12!=2^{10} \times 3^{5} \times 5^{2} \times 7 \times 11$.
Therefore, the number of divisors of 12! is $11 \times 6 \times 3 \times 2 \times 2$.
Since the divisor is a perfect square, the exponent in the prime factorization of the divisor must be even, and the divisor cannot be a multiple of 7 or 11.
Thus, the number of divisors that are perfect squares is $6 \times 3 \times 2$.
$$
\begin{array}{l}
\text { Hence } \frac{m}{n}=\frac{6 \times 3 \times 2}{11 \times 6 \times 3 \times 2 \times 2}=\frac{1}{22} \\
\Rightarrow m+n=1+22=23 .
\end{array}
$$