How many such pair of letters are there in the word ‘REVOLUTION’, each of which has as many letters between them in the word as they have in English alphabet (both in forward and backward direction)?

The correct answer is More than three.

To find the pairs of letters in the word REVOLUTION that have as many letters between them in the word as they have in the English alphabetical series (in both forward and backward directions), let us analyze the positions of each letter.

First, write down the alphabetical position (1–26) of each letter in REVOLUTION:
• R = 18
• E = 5
• V = 22
• O = 15
• L = 12
• U = 21
• T = 20
• I = 9
• O = 15
• N = 14

Now, let us check for valid letter pairs in both the forward and backward directions:

1. Forward Direction:
• R (18) to U (21): Counting forward from R: R(18), E(19), V(20), O(21). The 4th letter in position matches U(21). Thus, (R, U) is a valid pair.
• O (15) to T (20): Counting forward from the first O: O(15), L(16), U(17), T(18) — no match. But let us check O(15) to U(21): O(15), L(16), U(17) — no.
• L (12) to O (15): Counting forward from L: L(12), U(13), T(14), I(15). The letter at this position is I(9), not O. But L(12) to N(14): L(12), U(13), T(14), I(15), O(16), N(17) — no match.
• I (9) to N (14): Counting forward from I: I(9), O(10), N(11) — no match.

2. Backward Direction:
• N (14) to O (15): N and O are adjacent letters in the alphabet (14 and 15) and are adjacent in the word. Thus, (N, O) (reading right to left from N to the second O) is a valid pair.
• O (15) to T (20): Counting backward from N(14): N(14), O(15), I(16), T(17), U(18), L(19), O(20), V(21), E(22), R(23).
Let's check from I (9) going left:
- I(9), T(10), U(11), L(12)... L is in position 12, which matches! So (I, L) is a valid pair because between L and I in the word there are 2 letters (U, T), and in the English alphabet there are 2 letters (J, K) between I and L.
- From I(9): I(9), T(10), U(11), L(12), O(13), V(14), E(15), R(16)... no other match.
• T (20) to U (21): T and U are adjacent in the alphabet (20 and 21) and are adjacent in the word (reading from right to left: U, T). Thus, (T, U) is a valid pair.
• E (5) to I (9): Counting from E(5) to the right: E(5), V(6), O(7), L(8), U(9), T(10), I(11)... no match. Reading backwards from I(9) to E(5): I(9), T(8), U(7), L(6), O(5) — position matches O(15), not E.

Let's summarize the valid pairs found:
1. R - U (Forward: R_ _ _ U → R, E, V, O, U: 3 letters between them in word and in alphabet)
2. N - O (Backward: NO → 0 letters between them)
3. T - U (Backward: UT → 0 letters between them)
4. I - L (Backward: L U T I → 2 letters between them: U, T in word; J, K in alphabet)

Since we have found at least 4 valid pairs, the total number of such pairs is More than three.