Examine the statements: I. None but students are members of the club. (Only students can be members; hence every member is a student.) II. Some members of the club are married persons. III. All married persons are invited for the dance. Which conclusion follows?

Introduction / Context:This is a set-inclusion (syllogism) question. We are told who can be club members and who is invited for the dance. We must connect the sets correctly without over-generalising.

Given Data / Assumptions:

• “None but students are members” ⇒ Members ⊆ Students.
• “Some members are married” ⇒ There exists at least one member who is married; since Members ⊆ Students, some married members are married students.
• “All married persons are invited for the dance” ⇒ Married ⊆ Invited.

Concept / Approach:Chain subset relations: Members ⊆ Students; Married ⊆ Invited. Intersecting information (“some members are married”) yields “some married students exist” and those, being married, are invited.

Step-by-Step Solution:

Verification / Alternative check:Venn diagram: Draw Students; Members as a subset of Students; Married as an overlapping set. The overlap of Members and Married lies inside Students and is entirely inside Invited.

Why Other Options Are Wrong:

Common Pitfalls:Assuming converses or universalising “some” to “all.”

Final Answer:All married students of the club are invited for the dance.