Let the set $$S = \{2, 4, 8, 16, \ldots, 512\}$$ be partitioned into 3 sets $$A, B, C$$ with equal number of elements such that $$A \cup B \cup C = S$$ and $$A \cap B = B \cap C = A \cap C = \phi$$. The maximum number of such possible partitions of $$S$$ is equal to:

The set $$S = \{2, 4, 8, 16, 32, 64, 128, 256, 512\}$$ has 9 elements ($$2^1$$ to $$2^9$$).

We need to partition $$S$$ into 3 sets $$A, B, C$$ with equal number of elements, so each set has 3 elements.

Count total ways to partition 9 elements into 3 groups of 3.

The number of ways to choose 3 elements for set A from 9: $$\binom{9}{3}$$

Then choose 3 from remaining 6 for set B: $$\binom{6}{3}$$

The remaining 3 go to set C: $$\binom{3}{3}$$

Since the sets A, B, C are distinguishable (they are labeled), we don't divide by $$3!$$.

$$ \text{Total partitions} = \binom{9}{3} \times \binom{6}{3} \times \binom{3}{3} = 84 \times 20 \times 1 = 1680 $$

The correct answer is Option (1): 1680.