Question

Let the set $S = \{2, 4, 8, 16, ..., 512\}$ 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:

• A. 1680
• B. 1520
• C. 1710
• D. 1640

Answer 1

The Correct Option is A

The set \( S = \{ 2, 2^2, 2^3, \ldots, 2^9 \} \) contains 9 elements. To partition \( S \) into 3 subsets \( A, B, C \) of equal size, each subset must have exactly 3 elements.

The number of ways to partition the set can be calculated using the formula:

\[ \text{Number of partitions} = \frac{9!}{(3!3!3!)} \times 3!. \]

Expanding this expression:

\[ \text{Number of partitions} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4}{6 \times 6} \times 6 = 1680. \]

Therefore, the maximum number of such possible partitions of \( S \) is 1680.

Answer 2

Step 1: Understand the set S.
The given set is S = {2, 4, 8, 16, 32, 64, 128, 256, 512}.
It is a geometric progression with 9 elements, i.e., |S| = 9.

Step 2: Partition S into 3 sets with equal elements.
We are told that A, B, and C must each have an equal number of elements.
So, each set will have 9 / 3 = 3 elements.

Step 3: Count the number of ways to divide 9 elements into 3 groups of 3 elements each.
Number of ways to choose 3 elements for A = 9C3.
After that, 6 elements remain. Number of ways to choose 3 elements for B = 6C3.
The remaining 3 elements automatically go to C.

Total number of ways = (9C3 × 6C3 × 3C3) / 3!
We divide by 3! because the sets A, B, C are indistinguishable (their order doesn’t matter).

Step 4: Simplify the expression.
9C3 = 84, 6C3 = 20, and 3C3 = 1.
Total = (84 × 20 × 1) / 6 = 1680.

Final Answer: 1680