Laurie loves multiplying numbers in her head. One day she decides to multiply two $2$-digit numbers $x$ and $y$ such that $x\leq y$ and the two numbers collectively have at least three distinct digits. Unfortunately, she accidentally remembers the digits of each number in the opposite order (for example, instead of remembering $51$ she remembers $15$). Surprisingly, the product of the two numbers after flipping the digits is the same as the product of the two original numbers. How many possible pairs of numbers could Laurie have tried to multiply?
1. Let \( x = 10A + B \) and \( y = 10C + D \), where \( A, B, C, D \) are digits from 1 to 9 (since 0 would make the number not a 2-digit number).
2. According to the problem, Laurie remembers the digits in the opposite order, so she remembers \( x \) as \( 10B + A \) and \( y \) as \( 10D + C \).
3. The condition given is that the product of the original numbers is equal to the product of the numbers with flipped digits:
   \[
   (10A + B)(10C + D) = (10B + A)(10D + C)
   \]
4. Expanding both sides, we get:
   \[
   100AC + 10AD + 10BC + BD = 100BD + 10BC + 10AD + AC
   \]
5. Simplifying the equation by canceling out common terms on both sides:
   \[
   100AC + BD = 100BD + AC
   \]
6. Rearranging the terms, we get:
   \[
   99AC = 99BD
   \]
7. Dividing both sides by 99, we obtain:
   \[
   AC = BD
   \]
8. We need to find all combinations of \( A, B, C, D \) such that \( AC = BD \) and \( A, B, C, D \) are distinct digits.

9. Enumerate the possible values of \( A, B, C, D \) such that \( AC = BD \):
   - \( A = 1 \): \( C = 1, 2, 3, 4, 5, 6, 7, 8, 9 \)
   - \( A = 2 \): \( C = 1, 2, 3, 4 \)
   - \( A = 3 \): \( C = 1, 2, 3 \)
   - \( A = 4 \): \( C = 1, 2 \)
   - \( A = 5 \): \( C = 1, 2 \)
   - \( A = 6 \): \( C = 1 \)
   - \( A = 7 \): \( C = 1 \)
   - \( A = 8 \): \( C = 1 \)
   - \( A = 9 \): \( C = 1 \)

10. For each valid pair \((A, C)\), find corresponding pairs \((B, D)\) such that \( BD = AC \):
    - For \( A = 1 \), \( C = 1 \): \( B = 1, 2, 3, 4, 5, 6, 7, 8, 9 \), \( D = 1, 2, 3, 4, 5, 6, 7, 8, 9 \)
    - For \( A = 2 \), \( C = 1 \): \( B = 2, 4, 6, 8 \), \( D = 1, 2, 3, 4 \)
    - For \( A = 3 \), \( C = 1 \): \( B = 3, 6, 9 \), \( D = 1, 2, 3 \)
    - For \( A = 4 \), \( C = 1 \): \( B = 4, 8 \), \( D = 1, 2 \)
    - For \( A = 5 \), \( C = 1 \): \( B = 5, 10 \), \( D = 1, 2 \)
    - For \( A = 6 \), \( C = 1 \): \( B = 6 \), \( D = 1 \)
    - For \( A = 7 \), \( C = 1 \): \( B = 7 \), \( D = 1 \)
    - For \( A = 8 \), \( C = 1 \): \( B = 8 \), \( D = 1 \)
    - For \( A = 9 \), \( C = 1 \): \( B = 9 \), \( D = 1 \)

11. Count the number of valid pairs \((A, B, C, D)\) ensuring at least three distinct digits:
    - Each set of 4 digits contributes twice to the total number of pairs (since \((x = AB, y = CD)\) and \((x = BA, y = DC)\) are both valid).

12. Summing up all valid pairs, we get a total of 14 sets of 4 digits, each contributing twice:
    \[
    14 \times 2 = 28
    \]

The final answer is \(\boxed{28}\)