Key Concepts and Formulas

• Combinations: The number of ways to choose r objects from a set of n distinct objects is given by the binomial coefficient: ${n \choose r} = \frac{n!}{r!(n-r)!}$
• Partitioning a Set: Dividing a set into non-overlapping subsets.
• Permutations: The number of ways to arrange n distinct objects in a specific order is n!

Step-by-Step Solution

Step 1: Determine the number of elements in each set.

Since the set $S$ has 9 elements ($S = \{2, 4, 8, 16, 32, 64, 128, 256, 512\} = \{2^1, 2^2, 2^3, 2^4, 2^5, 2^6, 2^7, 2^8, 2^9\}$) and it is partitioned into 3 sets $A, B, C$ with an equal number of elements, each set must contain 3 elements.

Step 2: Choose the elements for set A.

We need to select 3 elements out of the 9 elements in $S$ to form set $A$. The number of ways to do this is given by the combination formula: ${9 \choose 3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$ There are 84 ways to choose the elements for set $A$.

Step 3: Choose the elements for set B.

After selecting the elements for set $A$, we are left with 6 elements in $S$. We need to select 3 elements out of these remaining 6 elements to form set $B$. The number of ways to do this is given by the combination formula: ${6 \choose 3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 5 \times 4 = 20$ There are 20 ways to choose the elements for set $B$.

Step 4: Choose the elements for set C.

After selecting the elements for sets $A$ and $B$, we are left with 3 elements in $S$. These remaining 3 elements will form set $C$. The number of ways to do this is given by the combination formula: ${3 \choose 3} = \frac{3!}{3!(3-3)!} = \frac{3!}{3!0!} = 1$ There is only 1 way to choose the elements for set $C$.

Step 5: Calculate the total number of partitions.

The total number of ways to partition the set $S$ into sets $A, B, C$ is the product of the number of ways to choose elements for each set. Therefore, the total number of partitions is: ${9 \choose 3} \cdot {6 \choose 3} \cdot {3 \choose 3} = 84 \times 20 \times 1 = 1680$

However, since the sets A, B, and C are indistinguishable (i.e., we don't care about the order in which we choose the sets), we must divide by the number of ways to arrange the three sets, which is $3! = 3 \times 2 \times 1 = 6$. Therefore, the number of such possible partitions is: $\frac{{9 \choose 3} \cdot {6 \choose 3} \cdot {3 \choose 3}}{3!} = \frac{1680}{6}$ This calculation is incorrect. The sets A, B, and C are distinguishable. Therefore, we DO NOT need to divide by 3!.

The number of partitions is: ${9 \choose 3} \cdot {6 \choose 3} \cdot {3 \choose 3} = 84 \times 20 \times 1 = 1680$

Common Mistakes & Tips

• Distinguishability of Sets: Carefully consider whether the sets in the partition are distinguishable or indistinguishable. If they are indistinguishable, you need to divide by the factorial of the number of sets. If they are distinguishable, you do not. In this case, the sets are distinguishable.
• Combination vs. Permutation: Remember to use combinations when the order of selection does not matter, and permutations when the order matters.
• Double-check calculations: Ensure your arithmetic is correct, especially when dealing with factorials and binomial coefficients.

Summary

The problem asks us to find the number of ways to partition a set of 9 elements into three sets of 3 elements each. We first calculate the number of ways to choose 3 elements for the first set, then the number of ways to choose 3 elements from the remaining elements for the second set, and finally, the remaining elements form the third set. Since the sets are distinguishable, we multiply the number of ways to choose each set to get the total number of partitions. The maximum number of such possible partitions is 1680.

Final Answer

The final answer is \boxed{1680}, which corresponds to option (D).