The set $$S = \{1, 2, 3, \ldots, 12\}$$ is to be partitioned into three sets $$A, B, C$$ of equal size. Thus, $$A \cup B \cup C = S, A \cap B = B \cap C = A \cap C = \phi$$. The number of ways to partition $$S$$ is

The three subsets $$A, B, C$$ are named, so they are regarded as distinct (labelled) groups.

Step 1: Choose the 4 elements that will go into set $$A$$.
Number of ways = $${}^{12}C_{4} = \frac{12!}{4!\,8!}$$.

Step 2: From the remaining 8 elements, choose the 4 elements for set $$B$$.
Number of ways = $${}^{8}C_{4} = \frac{8!}{4!\,4!}$$.

Step 3: The last 4 elements automatically form set $$C$$, so no further choice is needed.

Total number of partitions:
Multiply the counts from Steps 1 and 2:
$$ {}^{12}C_{4}\;{}^{8}C_{4} = \frac{12!}{4!\,8!}\;\frac{8!}{4!\,4!} = \frac{12!}{(4!)^3}. $$

Thus, the required number of ways to partition the set $$S$$ into three labelled subsets of equal size is $$\dfrac{12!}{(4!)^3}$$.

Option C which is: $$\frac{12!}{(4!)^3}$$