Syllogism — Which conclusion(s) necessarily follow? Statements: • All parrots are chicks. • All birds are chicks. Conclusions: I. Some birds are parrots. II. Some chicks are parrots.

Introduction / Context:
Two universal statements place both Parrots and Birds wholly inside Chicks. The trap is to infer an overlap between Birds and Parrots, which is not enforced by the premises.

Given Data / Assumptions:

• S1: Parrots ⊆ Chicks.
• S2: Birds ⊆ Chicks.
• Conclusions: I “Some Birds are Parrots.” II “Some Chicks are Parrots.”

Concept / Approach:
Two different subsets of the same superset need not intersect. Therefore C1 is not necessary. However, provided Parrots exist (standard assumption for such categories), then certainly some Chicks are Parrots, because all Parrots are Chicks.

Step-by-Step Solution:
1) From S1, if ∃x Parrot(x), then x is a Chick. This witnesses C2.2) For C1, we would need Parrots ∩ Birds ≠ ∅, which the premises do not guarantee. They could be disjoint subsets within Chicks.

Verification / Alternative check:
Diagram with Parrots and Birds as two non-overlapping circles inside Chicks: S1 and S2 both hold, C2 holds (pick any Parrot), while C1 fails.

Why Other Options Are Wrong:
Including C1 assumes an overlap not given; “none” ignores the existential that follows from S1 given non-emptiness of Parrots.

Common Pitfalls:
Believing subsets of the same superset must overlap (they need not).

Final Answer:
Only Conclusion II follows.