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.)

Correct — C. Rule: first letter to second is +3, second letter to third is +4.

DGK: D +3 gives G, G +4 gives K
HKO: H +3 gives K, K +4 gives O
MPT: M +3 gives P, P +4 gives T

ACG: A +2 gives C, then C +4 gives G. The first gap is +2, not +3, so ACG is the cluster that does not belong — option (c).

Why the others are wrong

• (a)
  DGK — D to G is +3 and G to K is +4, the shared pattern, so DGK sits inside the group and cannot be the outsider.
• (b)
  HKO — H to K is +3 and K to O is +4 — the same pair of gaps as DGK and MPT, so HKO belongs to the group.
• (d)
  MPT — M to P is +3 and P to T is +4, matching the other members, so MPT is not the odd one.

An odd-one-out on letter-clusters is settled by the internal gaps — the distance from each letter to the next inside the cluster — never by the letters themselves.

Convert to positions once (A = 1 up to Z = 26) and write the two gaps under each cluster. Three clusters will show the same pair of gaps and one will not.

The note under the question closes the usual escape route. The grouping is not about vowels or consonants, so counting them is wasted time.

Key facts

• DGK, HKO and MPT all run +3 then +4 between their letters.
• ACG runs +2 then +4, which is why it leaves the group.
• Positions used here: A=1, C=3, D=4, G=7, H=8, K=11, M=13, O=15, P=16, T=20.

Common traps

• Comparing the first letters of the four clusters instead of the gaps inside each one
• Counting vowels and consonants, which the question's own note rules out
• Reading A to C as +3 because A feels like position zero

No directly related past PYQ was found.