In the word CASTRAPHONE, how many pairs of letters are there such that the number of letters between them in the word is the same as the number of letters between them in the English alphabet sequence?

Introduction / Context:
This alphabet test question extends the same idea used in the previous problem. The word is CASTRAPHONE, and we are asked to find how many pairs of letters have the same number of letters between them in the word as in the alphabet. Because the word is longer, there are more possible pairs, and a systematic approach is essential to avoid missing any or counting extra pairs. The problem tests visual tracking, numerical counting, and familiarity with the alphabetical order.

Given Data / Assumptions:

• Word: C A S T R A P H O N E.
• We consider the standard English alphabet sequence.
• A pair consists of two letters in the word in their given order.
• For each pair, the number of letters between them in the word must equal the number of letters between them in the alphabet.
• We count each qualifying pair once.

Concept / Approach:
Because there are many possible pairs in a long word, it is best to write down the positions of letters in the word and note their alphabetical positions. For each pair, we calculate two numbers: the gap in the word and the gap in the alphabet. The gap in the word is the difference in positions minus one, and the gap in the alphabet is the difference in alphabetical positions minus one. If these two gaps match, that pair satisfies the condition. Using a structured scan ensures we do not overlook any valid pair.

Step-by-Step Solution:
Step 1: Write the letters of CASTRAPHONE with serial positions: C(1), A(2), S(3), T(4), R(5), A(6), P(7), H(8), O(9), N(10), E(11).Step 2: Note the alphabetical positions: C=3, A=1, S=19, T=20, R=18, P=16, H=8, O=15, N=14, E=5.Step 3: Examine relevant pairs and compare gaps. For example, pair S and T at positions 3 and 4 has zero letters between them in the word, and in the alphabet S and T are consecutive, so there are zero letters between them there as well. Therefore S–T is a valid pair.Step 4: Continue systematically and you will find six valid pairs: S–T, T–O, T–N, R–P, H–E, and O–N. In each of these cases, the number of letters between the two in the word matches the number of letters between them in alphabetical order.Step 5: For all other pairs, either the gap in the word and the gap in the alphabet do not match, or the letters are repeated in such a way that the counts do not line up. After checking all pairs, we confirm that there are exactly six qualifying pairs.

Verification / Alternative check:
A careful numerical check for each of the identified pairs confirms correctness. For example, T is the twentieth letter and O is the fifteenth, so they are separated by four letters in the alphabet (one less than the difference in their positions). In the word, T and O occur at positions 4 and 9, with four letters between them as well. Similar calculations verify that T–N, R–P, H–E, and O–N satisfy the condition. No additional pair satisfies both gap conditions, which confirms the count of six pairs.

Why Other Options Are Wrong:

• Options a, b, and c (3, 4, or 5 pairs) underestimate the number of qualifying pairs and typically result from incomplete checking or early stopping.
• Option e (2) is far too small and would mean that several valid pairs have been overlooked.

Common Pitfalls:
Because there are many possible pairs, students often lose track while counting and either skip some pairs or double count. Another common error is to forget to subtract one when counting letters between two positions in the word or in the alphabet, which leads to off by one mistakes. Writing positions clearly, proceeding in a fixed order, and comparing both word gaps and alphabet gaps carefully helps avoid such errors and leads to the correct answer of six pairs.

Final Answer:
There are 6 such pairs of letters in the word CASTRAPHONE.