Based on the English alphabetical order, three of the following four letter-clusters are alike in a certain way and thus form a group. Which is the letter-cluster that DOES NOT belong to that group? (Note: The odd man out is not based on the number of consonants/vowels or their position in the letter-cluster.)

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Accepted Answer

Let's analyze the letter clusters based on the alphabetical order and the difference between the positions of letters:

1. CHE: C(3), H(8), E(5)
   Differences: H - C = 8 - 3 = 5, E - H = 5 - 8 = -3

2. AGD: A(1), G(7), D(4)
   Differences: G - A = 7 - 1 = 6, D - G = 4 - 7 = -3

3. MSP: M(13), S(19), P(16)
   Differences: S - M = 19 - 13 = 6, P - S = 16 - 19 = -3

4. FLI: F(6), L(12), I(9)
   Differences: L - F = 12 - 6 = 6, I - L = 9 - 12 = -3

Observing the differences:
- For CHE, the first difference is 5, second is -3
- For AGD, MSP, and FLI, the first difference is 6, second is -3

Thus, AGD, MSP, and FLI share the same pattern of differences (6 and -3), while CHE differs with a first difference of 5.

Therefore, CHE is the odd one out. However, the question asks which letter-cluster does NOT belong to the group, and among the options, AGD is marked as correct in the options above. This is a mistake.

Re-evaluating:
- CHE: differences 5 and -3
- AGD: differences 6 and -3
- MSP: differences 6 and -3
- FLI: differences 6 and -3

So CHE is the odd one out.

Hence, the correct odd one out is CHE.

Correcting the options accordingly:
- CHE: isCorrect = true
- AGD, MSP, FLI: isCorrect = false

Final answer: CHE

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