Let the set S = {2, 4, 8, 16, ..., 512} be partitioned into 3 sets A, B, C with equal number of elem

Q: Wait, why do we treat the sets $A, B, C$ as labeled? The problem doesn't explicitly say they are distinct.

In set theory, when we define a partition into sets $A, B,$ and $C$, these names act as labels. If the problem asked to divide the set into 'three piles' without naming them, the sets would be unlabeled, and we would divide by $3!$ to account for the indistinguishable order of the piles. Since they are named $A, B,$ and $C$, the arrangement $(A=\{2,4,8\}, B=\{16,32,64\}, C=\{128,256,512\})$ is distinct from $(A=\{16,32,64\}, B=\{2,4,8\}, C=\{128,256,512\})$. Thus, they are labeled.

Q: Does the fact that the elements are powers of 2 change the combinatorics?

Not at all! This is a classic 'distractor' in JEE problems. Whether the elements are powers of 2, consecutive integers, or even random letters, the logic of partitioning a set of 9 distinct objects remains identical. The values themselves do not affect the number of ways to group them.

Q: Why can't we just use $9C3 \times 6C3 \times 3C3$?

Actually, you can! That is a perfectly valid way to think about it. $9C3$ chooses 3 elements for set $A$, then $6C3$ chooses 3 for set $B$, and $3C3$ chooses the remaining 3 for set $C$. If you calculate $\binom{9}{3} \times \binom{6}{3} \times \binom{3}{3}$, you get $\frac{9!}{3!6!} \times \frac{6!}{3!3!} \times 1 = \frac{9!}{3!3!3!}$, which is exactly the same result. Both methods lead to the same answer of 1680.

Question

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

Options:

(A) (1) 1680

(B) (2) 1640

(C) (3) 1520

(D) (4) 1710

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The set $S = \{2, 4, 8, 16, \dots, 512\}$ consists of powers of two, specifically $2^1, 2^2, 2^3, \dots, 2^9$. There are exactly $9$ distinct elements in this set.

In combinatorial terms, the specific values of these elements are merely labels. We are tasked with partitioning these $9$ distinct items into three distinct (labeled) containers: $A$, $B$, and $C$.

### The Logic of Labeled Containers

A partition requires that every element belongs to exactly one set, and the sets are mutually disjoint. Since we must distribute $9$ elements into $3$ sets such that each set has an equal number of elements, each set must contain exactly:

$$\frac{9}{3} = 3 	ext{ elements}$$

Because the sets $A$, $B$, and $C$ are labeled, the assignment of an element to a specific set matters. Placing an element into set $A$ is a distinct outcome from placing it into set $B$.

### The Multinomial Engine

To distribute $n$ distinct objects into $k$ labeled groups of sizes $n_1, n_2, \dots, n_k$, we utilize the **multinomial coefficient**. The formula is given by:

$$	ext{Number of ways} = \frac{n!}{n_1! \cdot n_2! \cdot \dots \cdot n_k!}$$

In this problem, $n = 9$ and the group sizes are $n_1 = 3, n_2 = 3,$ and $n_3 = 3$. Substituting these values, we obtain:

$$	ext{Number of ways} = \frac{9!}{3! \cdot 3! \cdot 3!}$$

This expression accounts for all permutations of the elements while correcting for the fact that the internal order of elements within each bucket is irrelevant.

### The Beauty of Cancellation

We now perform the arithmetic calculation:

$$\frac{9!}{3! \cdot 3! \cdot 3!} = \frac{9 	imes 8 	imes 7 	imes 6 	imes 5 	imes 4 	imes 3!}{3! 	imes (3 	imes 2 	imes 1) 	imes (3 	imes 2 	imes 1)}$$

Canceling one $3!$ from the numerator and denominator, we are left with:

$$\frac{9 	imes 8 	imes 7 	imes 6 	imes 5 	imes 4}{6 	imes 6}$$

Since $6 	imes 6 = 36$ and $9 	imes 4 = 36$, these terms cancel out entirely. The expression simplifies to:

$$8 	imes 7 	imes 6 	imes 5 = 1680$$

### Final Calculation

The total number of ways to partition the set $S$ into three equal, labeled sets is **1680**.

This result demonstrates how complex-looking problems can be reduced to fundamental combinatorial structures. By identifying the underlying multinomial distribution, we arrive at a clean and definitive solution.

Select Answer:

Visualized Solution

Analyzing Set $S$

Partitioning the Set

Equal Number of Elements

Grouping Distinct Objects

Applying the Formula

Expanding the Factorials

Simplifying the Expression

Final Calculation

The Sigma Insight: Formation of Groups

Analyzing the Setup

The Logic of Labeled Containers

The Multinomial Engine

The Beauty of Cancellation

Final Calculation

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Visualized Solution

Analyzing Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Partitioning the Set

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