Question

For $ n \geq 2 $, let $ S_n $ denote the set of all subsets of $ \{1, 2, 3, \ldots, n\} $ with no two consecutive numbers. For example, $ \{1, 3, 5\} \in S_6 $, but $ \{1, 2, 4\} \notin S_6 $. Then, find $ n(S_5) $.

Hint:

When solving problems involving subsets with restrictions (e.g., no consecutive elements), a recurrence relation is a useful approach. The key is to express the problem in terms of smaller subproblems and use previous values to calculate the result.

Answer 1

Correct Answer: 13

Determine the count of subsets of \( \{1, 2, 3, 4, 5\} \) that do not contain any consecutive elements.
Step 1: Problem Analysis
This problem is solvable using a recurrence relation. Define \( a_n \) as the number of valid subsets of \( \{1, 2, \ldots, n\} \) that exclude consecutive elements. 
Step 2: Recurrence Relation Derivation
For a set of size \( n \), consider two cases for the element \( n \):
1. If \( n \) is not in the subset, the problem reduces to finding valid subsets of \( \{1, 2, \ldots, n-1\} \), which is \( a_{n-1} \) ways.
2. If \( n \) is in the subset, then \( n-1 \) cannot be included. The problem then becomes finding valid subsets of \( \{1, 2, \ldots, n-2\} \), which is \( a_{n-2} \) ways.
The recurrence relation is therefore: \[ a_n = a_{n-1} + a_{n-2}. \] 
Step 3: Base Cases Identification
Establish the base cases:
For \( n=2 \), the set is \( \{1, 2\} \). Valid subsets are \( \emptyset, \{1\}, \{2\} \), so \( a_2 = 3 \).
For \( n=3 \), the set is \( \{1, 2, 3\} \). Valid subsets are \( \emptyset, \{1\}, \{2\}, \{1, 3\} \), so \( a_3 = 4 \). 
Step 4: Calculation of \( a_5 \)
Apply the recurrence relation to compute \( a_5 \):
\[ a_4 = a_3 + a_2 = 4 + 3 = 7, \]
\[ a_5 = a_4 + a_3 = 7 + 4 = 13. \] 
The total number of subsets of \( \{1, 2, 3, 4, 5\} \) without consecutive elements is: \[ n(S_5) = 13. \]