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

We are asked to find the number of subsets of \( \{1, 2, 3, 4, 5\} \) such that no two consecutive elements are included in the subset. 
Step 1: Analyzing the problem
This problem can be solved using a recurrence relation. We define \( a_n \) as the number of valid subsets of \( \{1, 2, \ldots, n\} \) where no two consecutive elements are selected. 
Step 2: Recurrence Relation
The recurrence relation can be described as follows: If \( n \) is not included in the subset, then we are simply choosing a subset from \( \{1, 2, \ldots, n-1\} \), which can be done in \( a_{n-1} \) ways. 
If \( n \) is included in the subset, then \( n-1 \) cannot be included, and we are choosing a subset from \( \{1, 2, \ldots, n-2\} \), which can be done in \( a_{n-2} \) ways. 
Thus, the recurrence relation is: \[ a_n = a_{n-1} + a_{n-2}. \] 
Step 3: Base Cases
We need the base cases: \( a_2 = 3 \), because the subsets of \( \{1, 2\} \) with no consecutive numbers are \( \emptyset, \{1\}, \{2\} \). \( a_3 = 4 \), because the subsets of \( \{1, 2, 3\} \) with no consecutive numbers are \( \emptyset, \{1\}, \{2\}, \{1, 3\} \). 
Step 4: Calculate \( a_5 \)
Now, we can use the recurrence relation to calculate \( a_5 \): \[ a_4 = a_3 + a_2 = 4 + 3 = 7, \] \[ a_5 = a_4 + a_3 = 7 + 4 = 13. \] 
Thus, the number of subsets of \( \{1, 2, 3, 4, 5\} \) with no two consecutive elements is: \[ n(S_5) = 13. \]