How many pairs of letters are there in the word ‘CREATIVE’ which have

How many pairs of letters are there in the word ‘CREATIVE’ which have as many letters between them in the word as in the alphabet?

1
2
3
4
This question was previously asked in
UPSC CISF-AC-EXE – 2019
The correct answer is (C) 3. We need to find pairs of letters in the word ‘CREATIVE’ that have the same number of letters between them in the word as they do in the English alphabet.
– List the letters in the word ‘CREATIVE’ with their positions in the word (1-indexed) and their positions in the alphabet (A=1, B=2, …):
C(1, 3), R(2, 18), E(3, 5), A(4, 1), T(5, 20), I(6, 9), V(7, 22), E(8, 5)
– A pair of letters (L1, L2) at word positions P1 and P2 forms a matching pair if the number of letters between them in the word, which is |P1 – P2| – 1, is equal to the number of letters between them in the alphabet, which is |AlphaPos(L1) – AlphaPos(L2)| – 1. This simplifies to checking if |P1 – P2| = |AlphaPos(L1) – AlphaPos(L2)|.
– Let’s check all pairs:
– C(1,3) with E(3,5): |1-3|=2, |3-5|=2. Match! (C-E)
– C(1,3) with E(8,5): |1-8|=7, |3-5|=2. No.
– E(3,5) with A(4,1): |3-4|=1, |5-1|=4. No.
– E(3,5) with I(6,9): |3-6|=3, |5-9|=4. No.
– E(3,5) with V(7,22): |3-7|=4, |5-22|=17. No.
– E(3,5) with E(8,5): |3-8|=5, |5-5|=0. No.
– A(4,1) with E(8,5): |4-8|=4, |1-5|=4. Match! (A-E)
– T(5,20) with V(7,22): |5-7|=2, |20-22|=2. Match! (T-V)
– Other pairs do not satisfy the condition. (Checked systematically in thought process).
– The three pairs are C-E (occurring forward from C to the first E), A-E (occurring forward from A to the second E), and T-V (occurring forward from T to V). The order in the alphabet doesn’t matter (|difference| is used).
– Note that C-E (positions 1 and 3 in word) have 1 letter (R) between them. In the alphabet, C and E have 1 letter (D) between them.
– A-E (positions 4 and 8 in word) have 3 letters (T, I, V) between them. In the alphabet, A and E have 3 letters (B, C, D) between them.
– T-V (positions 5 and 7 in word) have 1 letter (I) between them. In the alphabet, T and V have 1 letter (U) between them.
The question asks for the number of *pairs*, regardless of direction (forward or backward in the word). The method |P1 – P2| = |AlphaPos(L1) – AlphaPos(L2)| correctly accounts for this. We found 3 such pairs: C-E, A-E, and T-V.