To solve this problem, we need to determine the total number of unordered pairs of disjoint subsets of the set $ S = \{1, 2, 3, 4\} $.

Understanding the Problem:

Disjoint Subsets: Two subsets $ A $ and $ B $ are disjoint if they have no elements in common, i.e., $ A \cap B = \emptyset $.

Unordered Pairs: Pairs where $ \{A, B\} $ is considered the same as $ \{B, A\} $.

Approach:

Counting Ordered Pairs of Disjoint Subsets:

For each element in $ S $, it can be in:

Subset $ A $ only,

Subset $ B $ only,

Neither $ A $ nor $ B $.

Note: An element cannot be in both $ A $ and $ B $ because $ A $ and $ B $ are disjoint.

Therefore, each element has 3 choices.

Total number of ordered pairs $ (A, B) $ is $ 3^4 = 81 $.

Identifying Ordered Pairs where $ A = B $:

Since $ A $ and $ B $ are disjoint and $ A = B $, the only possibility is when both are the empty set.

So, there's 1 ordered pair where $ A = B = \emptyset $.

Calculating Unordered Pairs:

Ordered Pairs with $ A \ne B $: $ 81 - 1 = 80 $.

Unordered Pairs from Ordered Pairs with $ A \ne B $: Each unordered pair corresponds to 2 ordered pairs (since $ (A, B) $ and $ (B, A) $ are different but represent the same unordered pair).

Number of unordered pairs from $ A \ne B $ is $ \frac{80}{2} = 40 $.

Include the pair where $ A = B = \emptyset $: Add $ 1 $.

Total Unordered Pairs: $ 40 + 1 = 41 $.

Conclusion:

The total number of unordered pairs of disjoint subsets of $ S $ is 41.

Answer: Option D