In this alphabet-series odd-man-out question, select the odd group of letters from the given alternatives.

Introduction / Context:
This question deals with simple alphabet sequences. Each option is a group of three letters, and three options follow one descending pattern while one option follows a slightly different pattern. You must detect the underlying rule for how letters change from first to second and from second to third, and then identify which group does not fit that rule.

Given Data / Assumptions:

• Options: UTS, PNM, KIH and FDC.
• We use standard alphabet positions A = 1, B = 2, ..., Z = 26.
• We will analyse the step difference between consecutive letters in each group.

Concept / Approach:
All groups appear to show letters in decreasing order from left to right. However, the size of the step from one letter to the next may differ. A careful analysis of the numerical differences helps us find the single group that behaves differently. The main pattern that emerges is that in three of the groups the first and second letters differ by 2 positions, while in one group they differ by only 1 position.

Step-by-Step Solution:
Step 1: Convert UTS to numeric positions. U = 21, T = 20 and S = 19. The differences are T - U = 20 - 21 = -1 and S - T = 19 - 20 = -1. So in UTS, the letters decrease by 1 and then by 1 again. Step 2: Convert PNM to positions. P = 16, N = 14 and M = 13. The differences are N - P = 14 - 16 = -2 and M - N = 13 - 14 = -1. The first step is -2, the second step is -1. Step 3: Convert KIH to positions. K = 11, I = 9 and H = 8. The differences are I - K = 9 - 11 = -2 and H - I = 8 - 9 = -1. Again we see a -2 step followed by a -1 step. Step 4: Convert FDC to positions. F = 6, D = 4 and C = 3. The differences are D - F = 4 - 6 = -2 and C - D = 3 - 4 = -1. So this group also shows -2 then -1. Step 5: Compare the patterns. PNM, KIH and FDC all follow the pattern of a drop of 2 positions from the first letter to the second, followed by a drop of 1 position from the second to the third. UTS is the only group where both drops are -1. Step 6: Conclude that UTS is the odd group because its step pattern is (-1, -1) instead of (-2, -1).

Verification / Alternative check:
If we try to classify based purely on whether letters are descending, then all four groups would appear similar, which does not help. The examiner clearly intends a more specific rule about how many positions each letter moves. The fact that three groups share (-2, -1) exactly and one group does not gives a very strong basis for classification. No other simple pattern cleanly isolates just one group, so our conclusion is reliable.

Why Other Options Are Wrong:
PNM is not odd because it follows the (-2, -1) pattern in its letter steps. KIH also follows (-2, -1) and aligns with PNM and FDC. FDC follows the same pattern as well. Since these three groups share identical step sizes, they form the main category and cannot be considered the odd-man-out. Only UTS shows a different pattern of (-1, -1) and therefore stands out.

Common Pitfalls:
Some learners focus only on whether the letters are consecutive or not and might mistakenly think that PNM, KIH and FDC are irregular because they skip one letter between the first and second positions. However, the aim in an odd-man-out question is not to find groups that look strange individually, but to find a single group that does not share a common rule. If you compute and compare the exact step sizes, the pattern becomes clear and UTS emerges as the only exception.

Final Answer:
The odd group of letters is UTS, because its letters decrease by 1 step twice, whereas in the other groups the letters first decrease by 2 steps and then by 1 step.