(3) Six families are sequentially numbered as $1, 2, 3, 4, 5, 6$. Each family has three members, and they all gather to play a game. The game is played in groups of three, one after another. The probability that members of the same group come from different families is ( ).
(A) $\frac{5}{68}$
(B) $\frac{15}{68}$
(C) $\frac{45}{68}$
(D) $\frac{5}{204}$
(3) From 18 people, selecting 3 people, the number of different ways to select is $\mathrm{C}_{18}^{3}=\frac{18 \times 17 \times 16}{3 \times 2 \times 1}=816$.

Since the three people in a family have the same number, to ensure that the selected members have different numbers, one should first select 3 numbers from 6 numbers, which has $\mathrm{C}_{6}^{3}$ ways. For each number, there are 3 ways to select a person. Therefore, the number of ways to select group members from different families is
$$
\mathrm{C}_{6}^{3} \cdot 3^{3}=\frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times 3^{3}=5 \times 4 \times 3^{3}=540 .
$$

Thus, the probability that the three members of a group come from different families is
$$
\frac{540}{816}=\frac{5 \times 4 \times 3^{3}}{3 \times 17 \times 16}=\frac{5 \times 3^{2}}{17 \times 4}=\frac{45}{68} .
$$

Therefore, the answer is C.