In the number $48927269485$, how many pairs of digits have the same number of digits between them as in the original number, considering both forward and backward directions?

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**Solution:**

1. Write down the number with indices for clarity:

Number: 4 8 9 2 7 2 6 9 4 8 5

Indices: 1 2 3 4 5 6 7 8 9 10 11

2. We need to find pairs of digits that are the same and check if the count of digits between them matches the count of digits between the same digits in the original number.

3. Identify pairs of identical digits:
   - Digit 4 appears at positions 1 and 9
   - Digit 8 appears at positions 2 and 10
   - Digit 9 appears at positions 3 and 8
   - Digit 2 appears at positions 4 and 6

4. Calculate the number of digits between each pair:
   - For 4: positions 1 and 9 → digits between = 9 - 1 - 1 = 7
   - For 8: positions 2 and 10 → digits between = 10 - 2 - 1 = 7
   - For 9: positions 3 and 8 → digits between = 8 - 3 - 1 = 4
   - For 2: positions 4 and 6 → digits between = 6 - 4 - 1 = 1

5. Now, check if the pairs have the same number of digits between them both forward and backward:
   - For 4 and 9, the number of digits between is 7 and 4 respectively, so they differ.
   - For 8 and 10, the digits between is 7.
   - For 9 and 8, digits between is 4.
   - For 2 and 6, digits between is 1.

6. The pairs (4,9), (8,10), (9,8), and (4,9) satisfy the condition of having the same number of digits between them in both directions.

7. Hence, there are **four** such pairs.

**Answer:** Four

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