Among $3n$ cards labeled with the numbers $1,2,3 , \cdots , 3 n$ ($n$ is a natural number), two cards are randomly drawn, and the two numbers on them are denoted as $a , b$ ($a < b$) respectively. Let $\mathrm { P } _ { n }$ be the probability that $3 a < b$. The following is the process of finding $\lim _ { n \rightarrow \infty } \mathrm { P } _ { n }$.\\
The number of ways to draw 2 cards from $3 n$ cards is ${ } _ { 3 n } \mathrm { C } _ { 2 }$.\\
When $3 a < b$, we have $b \leqq 3 n$, so $1 \leqq a < n$.\\
Therefore, if $a = k$, the number of cases of $b$ satisfying $3 a < b$ is (A), so
$$\mathrm { P } _ { n } = \frac { \text { (B) } } { { } _ { 3 n } \mathrm { C } _ { 2 } } \text { . }$$
Therefore, $\lim _ { n \rightarrow \infty } \mathrm { P } _ { n } =$ (C).

What are the correct values for (A), (B), and (C)? [4 points]\\
\begin{tabular}{lll}
 & (A) & (B) & (C) \\
(1) & $3 ( n - k )$ & $\frac { 3 } { 2 } n ( n - 1 )$ & $\frac { 1 } { 3 }$ \\
(2) & $3 ( n - k )$ & $\frac { 3 } { 2 } n ( n - 1 )$ & $\frac { 2 } { 3 }$ \\
(3) & $3 ( n - k )$ & $3 n ( n - 1 )$ & $\frac { 2 } { 3 }$ \\
(4) & $3 ( n - k + 1 )$ & $3 n ( n - 1 )$ & $\frac { 1 } { 3 }$ \\
(5) & $3 ( n - k + 1 )$ & $3 n ( n - 1 )$ & $\frac { 2 } { 3 }$ \\
\end{tabular}