How many such pairs of digits are there in the number ‘95126139', which have as many digits between them in the number (both forward and backward direction) as they have between them in the Numeric Series?

The correct option is Four.

To find the pairs of digits in the given number '95126139' that have as many digits between them as in the standard numeric series (counting in both forward and backward directions), let us examine the number position by position.

The given number is 95126139.

Let us list the positions of each digit:

Position 1: 9
Position 2: 5
Position 3: 1
Position 4: 2
Position 5: 6
Position 6: 1
Position 7: 3
Position 8: 9

Now, let us check pairs in the forward direction:

1. 1 and 2: From Position 3 (digit 1) to Position 4 (digit 2), there are 0 digits between them in the number, and there are 0 digits between 1 and 2 in the numeric series. Thus, (1, 2) forms a valid pair.
2. 1 and 3: From Position 3 (digit 1) to Position 7 (digit 3), there are 3 digits (2, 6, 1) between them in the number. In the numeric series, the number of digits between 1 and 3 is 1 (which is 2). Since 3 ≠ 1, this is not a pair.
3. 1 and 3: From Position 6 (digit 1) to Position 7 (digit 3), there are 0 digits between them in the number. In the numeric series, the difference is $3-1=2$ (so 1 digit between them). Not a pair.
4. 5 and 6: From Position 2 (digit 5) to Position 5 (digit 6), there are 2 digits (1, 2) between them in the number. In the numeric series, there are 0 digits between 5 and 6. Not a pair.
5. 2 and 3: From Position 4 (digit 2) to Position 7 (digit 3), there are 2 digits (6, 1) between them in the number. In the numeric series, there are 0 digits between 2 and 3. Not a pair.

Now, let us check pairs in the backward direction:

1. 3 and 1: From Position 7 (digit 3) to Position 3 (digit 1), going backward, the distance is $7-3=4$, so there are 3 digits between them (6, 2, 1). In the numeric series, between 1 and 3 there is 1 digit. Not a pair.
2. 3 and 1: From Position 7 (digit 3) to Position 6 (digit 1), going backward, there are 0 digits between them. In the numeric series, between 1 and 3 there is 1 digit. Not a pair.
3. 3 and 2: From Position 7 (digit 3) to Position 4 (digit 2), going backward, there are 2 digits (1, 6) between them in the number. But in the numeric series, the difference between 3 and 2 is 1 (0 digits between them).
4. 3 and 5: From Position 7 (digit 3) to Position 2 (digit 5), going backward, there are 4 digits (1, 6, 2, 1) between them. In the numeric series, the difference between 5 and 3 is 2, so there is 1 digit (4) between them. Not a pair.
5. 6 and 5: From Position 5 (digit 6) to Position 2 (digit 5), going backward, there are 2 digits (2, 1) between them in the number. In the numeric series, the digits 5 and 6 have 0 digits between them.
6. 6 and 9: From Position 5 (digit 6) to Position 1 (digit 9), going backward, there are 3 digits (2, 1, 5) between them. In the numeric series, between 6 and 9 there are 2 digits (7, 8). Not a pair.
7. Let us re-count carefully systematically:

Let us count all pairs where $|{\text{pos}}_2-{\text{pos}}_1|=|{\text{digit}}_2-{\text{digit}}_1|$:

1. 1 (pos 3) and 2 (pos 4): $|4-3|=1$ and $|2-1|=1$. (Pair 1: 1-2)
2. 5 (pos 2) and 1 (pos 6): $|6-2|=4$ and $|5-1|=4$. (Pair 2: 5-1)
3. 5 (pos 2) and 1 (pos 3): $|3-2|=1$ and $|5-1|=4$ (No).
4. 6 (pos 5) and 3 (pos 7): $|7-5|=2$ and $|6-3|=3$ (No).
5. 1 (pos 3) and 6 (pos 5): $|5-3|=2$ and $|6-1|=5$ (No).
6. 2 (pos 4) and 6 (pos 5): $|5-4|=1$ and $|6-2|=4$ (No).
7. 6 (pos 5) and 9 (pos 8): $|8-5|=3$ and $|9-6|=3$. (Pair 3: 6-9)
8. 5 (pos 2) and 9 (pos 8): $|8-2|=6$ and $|9-5|=4$ (No).
9. 1 (pos 3) and 3 (pos 7): $|7-3|=4$ and $|3-1|=2$ (No).
10. 1 (pos 6) and 3 (pos 7): $|7-6|=1$ and $|3-1|=2$ (No).
11. 9 (pos 1) and 5 (pos 2): $|2-1|=1$ and $|9-5|=4$ (No).
12. 9 (pos 1) and 6 (pos 5): $|5-1|=4$ and $|9-6|=3$ (No).
13. 2 (pos 4) and 3 (pos 7): $|7-4|=3$ and $|3-2|=1$ (No).
14. 5 (pos 2) and 2 (pos 4): $|4-2|=2$ and $|5-2|=3$ (No).
15. 2 (pos 4) and 1 (pos 6): $|6-4|=2$ and $|2-1|=1$ (No).
16. 5 (pos 2) and 3 (pos 7): $|7-2|=5$ and $|5-3|=2$ (No).
17. 1 (pos 3) and 9 (pos 8): $|8-3|=5$ and $|9-1|=8$ (No).
18. 5 (pos 2) and 1 (pos 3): $|3-2|=1$ and $|5-1|=4$ (No).
19. 1 (pos 6) and 9 (pos 8): $|8-6|=2$ and $|9-1|=8$ (No).
20. 3 (pos 7) and 9 (pos 8): $|8-7|=1$ and $|9-3|=6$ (No).
21. 9 (pos 1) and 1 (pos 3): $|3-1|=2$ and $|9-1|=8$ (No).
22. 9 (pos 1) and 1 (pos 6): $|6-1|=5$ and $|9-1|=8$ (No).
23. 9 (pos 1) and 3 (pos 7): $|7-1|=6$ and $|9-3|=6$. (Pair 4: 9-3)

Thus, we have identified all 4 valid pairs:

1. (1, 2) between Position 3 (1) and Position 4 (2): $|4-3|=1=|2-1|$
2. (5, 1) between Position 2 (5) and Position 6 (1): $|6-2|=4=|5-1|$
3. (6, 9) between Position 5 (6) and Position 8 (9): $|8-5|=3=|9-6|$
4. (9, 3) between Position 1 (9) and Position 7 (3): $|7-1|=6=|9-3|$

Therefore, there are exactly Four such pairs of digits in the number.