Six members of a panel, A, B, C, D, E and F, are sitting in a row. A is to the right of C but to the left of B. E is sitting to the left of C and to the right of D, and D is sitting to the right of F. Based on this seating arrangement, which two members are sitting at the two ends of the row?

Introduction / Context:
This is a linear seating arrangement problem involving six panel members. You are given a set of positional relationships among A, B, C, D, E and F sitting in a single row. The question asks which two members occupy the two ends of the row. These puzzles are standard in verbal reasoning and require converting verbal clues into an ordered sequence.

Given Data / Assumptions:
- Members: A, B, C, D, E and F. - All six are sitting in a straight row with consecutive seats. - A is to the right of C but to the left of B, so C, A, B appear in that order from left to right. - E is left of C and right of D, so D, E, C appear in that order from left to right. - D is to the right of F, so F appears to the left of D. - We must identify the two members sitting at the extreme left and extreme right ends.

Concept / Approach:
The approach is to build a single consistent order that satisfies all constraints simultaneously. We start by forming smaller ordered segments like C, A, B and D, E, C. Then we place these segments relative to F using the condition that F is to the left of D. Because all six members must fit into one row, the constraints will force a unique sequence. The members at the two ends of this sequence are the answer.

Step-by-Step Solution:
Step 1: From A is to the right of C but left of B, we obtain the ordered segment C, A, B from left to right. Step 2: From E is left of C and right of D, we obtain the segment D, E, C from left to right. Step 3: Combining the segments D, E, C and C, A, B gives the longer ordered chain D, E, C, A, B. Step 4: The remaining member is F. From D is to the right of F, we know F must sit somewhere to the left of D. Step 5: Therefore, F must come before D in the overall left to right order, giving F, D, E, C, A, B. Step 6: Now we have a complete, consistent sequence of all six members: F at the extreme left end and B at the extreme right end.

Verification / Alternative check:
Check each original condition against the derived order F, D, E, C, A, B. A is to the right of C and to the left of B, which is satisfied as C, A, B appear in that order. E is to the left of C and to the right of D, which holds as D, E, C occur consecutively. D is to the right of F, which is satisfied since F is at the extreme left. There is no way to place any member differently without violating at least one condition, so this sequence is unique and correct.

Why Other Options Are Wrong:
- Options involving D at an end, such as D, F or D, B, contradict the condition that D must have F to its left. - Option A, B suggests A and B at the ends, but this ignores the members D, E, C and F and does not satisfy all relative positions. - Option C, F places C at an end, which conflicts with its required neighbors E and A in the center of the row. Only F, B correctly represent the extreme left and right members of the valid arrangement.

Common Pitfalls:
A common mistake is to treat each condition in isolation and not merge them into a single coherent order. Some students also misinterpret to the right of and to the left of as meaning immediate neighbors, which is not stated here. Others may forget to include F in the arrangement correctly. Drawing the row explicitly and progressively inserting each member according to the clues minimizes such mistakes.

Final Answer:
The two members sitting at the ends of the row are F at the extreme left and B at the extreme right.