Let $A$ and $B$ be two non-empty subsets of $X = \{1, 2, . . . , 11 \}$ with $A \cup B = X$. Let
$P_A$ be the product of all elements of $A$ and let $P_B$ be the product of all elements of $B$.
Find the minimum and maximum possible value of $P_A +P_B$ and find all possible equality
cases.

[i]Proposed by Greece[/i]
1. **Understanding the Problem:**
   We are given two non-empty subsets \( A \) and \( B \) of \( X = \{1, 2, \ldots, 11\} \) such that \( A \cup B = X \). We need to find the minimum and maximum possible values of \( P_A + P_B \), where \( P_A \) is the product of all elements in \( A \) and \( P_B \) is the product of all elements in \( B \).

2. **Maximum Value of \( P_A + P_B \):**
   - Since \( A \cup B = X \), the product of all elements in \( X \) is \( 11! \).
   - If \( A \) and \( B \) are such that \( A = X \) and \( B = X \), then \( P_A = P_B = 11! \).
   - Therefore, \( P_A + P_B = 11! + 11! = 2 \times 11! \).
   - Equality occurs when \( A = X \) and \( B = X \), or \( A = X \setminus \{1\} \) and \( B = X \), or \( A = X \) and \( B = X \setminus \{1\} \).

3. **Minimum Value of \( P_A + P_B \):**
   - We need to minimize \( P_A + P_B \) under the constraint \( P_A \times P_B = 11! \).
   - If there exists some \( 2 \leq c \leq 11 \) such that \( c \in A \) and \( c \in B \), then for the set \( A' = A \setminus \{c\} \), we would have \( A' \cup B = X \) and \( P_{A'} + P_B = \frac{1}{c} P_A + P_B < P_A + P_B \). Thus, \( A \cap B \) can only contain \( 1 \).
   - We need to find \( P_A \) and \( P_B \) such that \( P_A \times P_B = 11! \) and \( |P_A - P_B| \) is minimized.

4. **Calculating \( \sqrt{11!} \):**
   - \( 11! = 2^8 \times 3^4 \times 5^2 \times 7 \times 11 \).
   - \( \sqrt{11!} = 2^4 \times 3^2 \times 5 \times \sqrt{77} = 720 \sqrt{77} \).
   - Approximating \( \sqrt{77} \), we get \( 8.774 < \sqrt{77} < 8.775 \).
   - Therefore, \( 6317 < 720 \sqrt{77} < 6318 \).

5. **Finding the Nearest Number to \( 6317 \) Divisible by \( 11 \):**
   - We check numbers around \( \frac{6317}{11} \approx 574 \):
     - \( 567 = 3^4 \times 7 \)
     - \( 576 = 2^6 \times 3^2 \)
   - We check the sums:
     - \( 3^4 \times 7 \times 11 + 2^8 \times 5^2 = 6237 + 6400 = 12637 \)
     - \( 2^6 \times 3^2 \times 11 + 2^2 \times 3^2 \times 5^2 \times 7 = 6336 + 6300 = 12636 \)

6. **Conclusion:**
   - The minimum value of \( P_A + P_B \) is \( 12636 \).
   - The maximum value of \( P_A + P_B \) is \( 2 \times 11! \).

The final answer is \( \boxed{ 12636 \leq P_A + P_B \leq 2 \times 11! } \)