`X={1,2,3,....2017}` and `AsubX; BsubX;AuuBsubX` here `PsubQ` denotes that P is subset of `Q(P!=Q)` Then number of ways of selecting unordered pair of sets A and B such that `AuuBsubX.`

Options

A. `((4^(2017)-3^(2017))+(2^(2017)-1))/(2)`
B. `((4^(2017)-3^(2017)))/(2)`
C. `(4^(2017)-3^(2017)+2^(2017))/(2)`
D. none

Answer

To solve the problem of finding the number of unordered pairs of sets \( A \) and \( B \) such that \( A \cup B \subseteq X \) where \( X = \{1, 2, 3, \ldots, 2017\} \), we can follow these steps:

### Step 1: Understand the Problem
We need to find the number of unordered pairs of subsets \( (A, B) \) of \( X \) such that \( A \cup B \) is also a subset of \( X \). The condition \( A \neq B \) must also hold.

### Step 2: Determine the Choices for Each Element
For each element \( x \in X \), there are four possible choices regarding its membership in sets \( A \) and \( B \):
1. \( x \) is in \( A \) but not in \( B \).
2. \( x \) is in \( B \) but not in \( A \).
3. \( x \) is in both \( A \) and \( B \).
4. \( x \) is in neither \( A \) nor \( B \).

### Step 3: Total Combinations
Since there are 2017 elements in \( X \) and each element has 4 independent choices, the total number of ways to choose \( A \) and \( B \) is:
\[ 4^{2017} \]

### Step 4: Exclude the Invalid Case
However, we need to exclude the cases where \( A = B \). If \( A = B \), then for each element, we have only 2 choices:
1. The element is in both \( A \) and \( B \).
2. The element is in neither \( A \) nor \( B \).

Thus, the number of ways to choose \( A \) and \( B \) such that \( A = B \) is:
\[ 2^{2017} \]

### Step 5: Calculate Valid Combinations
To find the number of unordered pairs \( (A, B) \) such that \( A \cup B \subseteq X \) and \( A \neq B \), we subtract the invalid cases from the total combinations:
\[ \text{Valid pairs} = 4^{2017} - 2^{2017} \]

### Step 6: Adjust for Unordered Pairs
Since we are looking for unordered pairs, we need to divide the result by 2. However, we must also account for the case where \( A \) and \( B \) are equal, which we have already excluded:
\[ \text{Unordered pairs} = \frac{4^{2017} - 2^{2017}}{2} \]

### Final Result
Thus, the final number of unordered pairs of sets \( A \) and \( B \) such that \( A \cup B \subseteq X \) is:
\[ \frac{4^{2017} - 2^{2017}}{2} \]