Question

How many pairs of sets (S,T) are possible among the subsets of 1, 2, 3, 4, 5, 6 that satisfy the condition that S is a subset of T? 
 

• A. 729
• B. 728
• C. 665
• D. 664

Hint:

For problems of the form "Find the number of pairs (S,T) such that \(S \subseteq T \subseteq A\)", where A has \(n\) elements, the answer is always \(3^n\). This is a useful formula to remember for competitive exams.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Let the universal set be \(A = \{1, 2, 3, 4, 5, 6\}\). We need to find the number of ordered pairs of sets \((S, T)\) such that \(S \subseteq T \subseteq A\). This means S is a subset of T, and T itself is a subset of A.
Step 2: Key Formula or Approach:
Consider each element of the set A individually. For any element \(x \in A\), there are three distinct possibilities with respect to the sets S and T, given the condition \(S \subseteq T\):
1. \(x\) is not in T (and therefore, it cannot be in S). In set notation: \(x \notin T\) (which implies \(x \notin S\)).
2. \(x\) is in T, but not in S. In set notation: \(x \in T\) and \(x \notin S\).
3. \(x\) is in T, and it is also in S. In set notation: \(x \in T\) and \(x \in S\).
The case where \(x \in S\) but \(x \notin T\) is not allowed because it violates the condition \(S \subseteq T\).
Step 3: Detailed Calculation:
1. The universal set A has 6 elements.
2. For each of these 6 elements, we have 3 independent choices as described above.
- For element '1', there are 3 possibilities.
- For element '2', there are 3 possibilities.
- For element '3', there are 3 possibilities.
- For element '4', there are 3 possibilities.
- For element '5', there are 3 possibilities.
- For element '6', there are 3 possibilities.
3. Since the choice for each element is independent of the others, the total number of ways to form the pair of sets (S, T) is the product of the number of choices for each element.
\[ \text{Total number of pairs} = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6 \] 4. Calculate the value of \(3^6\):
\[ 3^6 = (3^3)^2 = 27^2 = 729 \] Step 4: Final Answer:
There are 729 possible pairs of sets (S,T).
Step 5: Why This is Correct:
The element-wise approach correctly accounts for all possible combinations. By considering the "location" of each of the 6 elements (either outside T, inside T but outside S, or inside S), we cover all valid set constructions. The multiplication principle applies because the placement of each element is an independent event.