12) The answer is $(\mathbf{A})$. To simplify the problem, one can think that the boys start calling their respective girlfriends one after the other. The order in this case is irrelevant. The first one chooses any frequency and finds the line free. The second one then chooses a frequency at random; regardless of the frequency occupied by the first boy, the probability that the second one chooses a different frequency (and thus finds the line free) is $\frac{3}{4}$. When the third one tries, if the first two frequencies are already occupied, the probability of choosing one of the two still free frequencies is $\frac{2}{4}=\frac{1}{2}$. Finally, for the fourth one, only one frequency will remain free, so the probability for him to occupy the only remaining free frequency is $\frac{1}{4}$. The probability that all three of these independent cases occur is therefore $\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{1}{4}=\frac{3}{32}$.

SECOND SOLUTION.

The random choice of one frequency out of 4, by 4 users, can happen in $4^{4}=256$ ways; among these choices, those in which the occupied lines are all different are $4 !=24$. The probability of falling into one of these cases is $\frac{24}{256}=\frac{3}{32}$.
12) Four boys want to call their respective girlfriends at the same time. Each phone can operate on four distinct frequencies. If two phones activate on the same frequency, the communication drops. If each boy does not know which frequency the other three choose, what is the probability that all four can talk to their girlfriends?
(A) $\frac{3}{32}$
(B) $\frac{3}{64}$
(C) $\frac{1}{256}$
(D) $\frac{1}{16}$
(E) $\frac{9}{128}$.