How many letter pairs exist in the term JOURNEY which contain the same number of letters between them in the given word as in the English alphabetical series (counting both forward and backward)?

The correct option is Three.

To find the number of letter pairs in the word JOURNEY that contain the same number of letters between them as in the English alphabetical series (counting in both forward and backward directions), let's first list the position of each letter in the English alphabet:

J = 10
O = 15
U = 21
R = 18
N = 14
E = 5
Y = 25

Now, let's examine the word letter-by-letter in both directions:

1. Forward Direction:

- J and N: In the word, there are 3 letters (O, U, R) between J (position 1) and N (position 5). In the alphabetical series, there are also 3 letters (K, L, M) between J (10) and N (14). Hence, (J, N) is a valid pair.
- U and Y: In the word, there are 3 letters (R, N, E) between U (position 3) and Y (position 7). In the alphabetical series, there are also 3 letters (V, W, X) between U (21) and Y (25). Hence, (U, Y) is a valid pair.

2. Backward Direction:

- E and J: Counting backwards from E (position 6) to J (position 1), there are 4 letters (N, R, U, O) between them in the word. In the alphabetical series, there are also 4 letters (F, G, H, I) between E (5) and J (10). Hence, (E, J) is a valid pair.

We can also verify this mathematically by checking if the absolute difference in word indices equals the absolute difference in alphabetical positions:

- For J and N: $|1-5|=4$ and $|10-14|=4$
- For U and Y: $|3-7|=4$ and $|21-25|=4$
- For E and J: $|6-1|=5$ and $|5-10|=5$

Therefore, there are exactly 3 such letter pairs: J - N, U - Y, and E - J.