Classroom with 5 rows: A, B, C, D, E are each in a different row, front (1st) to back (5th). A sits behind C but in front of B; C sits behind E; D sits in front of E. What is the order from the first row to the last?

Introduction / Context:
This is a linear ordering problem across five rows, from front (first row) to back (last row). We must respect several “in front of / behind” constraints.

Given Data / Assumptions:

• A is behind C but in front of B ⇒ C < A < B (where “<” means nearer the front).
• C is behind E ⇒ E < C.
• D is in front of E ⇒ D < E.
• All five are in distinct rows 1–5.

Concept / Approach:
Translate each phrase to an inequality chain from front (smaller index) to back (larger index) and then merge chains into a single total order.

Step-by-Step Solution:
From D < E and E < C we get D < E < C.From C < A < B we append to get D < E < C < A < B.Mapping to rows 1→5 yields: D (1st), E (2nd), C (3rd), A (4th), B (5th).

Verification / Alternative check:
Each original statement is satisfied by DECAB front to back.

Why Other Options Are Wrong:
They break at least one inequality (e.g., placing A ahead of C or D behind E).

Common Pitfalls:
Confusing “front” vs “back” ordering; we consistently treat “in front” as a smaller index.

Final Answer:
DECAB