Syllogism — Cars, Jeeps, Boxes, and Pens Statements: • Some cars are jeeps. • All the boxes are jeeps. • All the pens are cars. Conclusions: (1) Some cars are boxes. (2) No pen is a jeep. (3) Some boxes are cars.

Introduction / Context:
This question mixes “some” with two universal statements. The danger is to infer overlaps between Cars and Boxes or Pens and Jeeps without explicit compulsion.

Given Data / Assumptions:

• Some Cars (C) are Jeeps (J): C∩J ≠ ∅.
• All Boxes (B) are Jeeps (J): B ⊆ J.
• All Pens (P) are Cars (C): P ⊆ C.
• Conclusions: (1) Some C are B. (2) No P is J. (3) Some B are C.

Concept / Approach:
To prove “some X are Y,” we need a compelled intersection. To prove “No X is Y,” we need a compelled disjointness. Here, neither is forced between Cars and Boxes nor between Pens and Jeeps.

Step-by-Step Solution:

Verification / Alternative check:
Draw J as a set. Place B entirely inside J but outside C. Place C partially overlapping J (to satisfy “some C are J”) and put P inside the C area that is outside J. All premises hold; (1), (2), and (3) are false simultaneously, proving that none of the three conclusions must follow.

Why Other Options Are Wrong:

• Pairs like (1)&(2) or (1)&(3) assume overlaps not forced by the statements.
• (2)&(3) also fails because (2) is not compelled and (3) is not compelled either.

Common Pitfalls:
Assuming that two sets contained in a third must overlap each other. Subsets of a common superset can still be disjoint.

Final Answer:
None of three