From a pool of 8 ladies and 7 gentlemen we must form a committee of 3 ladies and 4 gentlemen. Mrs. X refuses to serve if Mr. Y is a member of the same committee. How many committees are possible under this restriction?

Introduction / Context:This is a combinations problem with an exclusion constraint disallowing the simultaneous presence of two specific members (Mrs. X and Mr. Y) on the same committee.

Given Data / Assumptions:

• Total ladies = 8; total gentlemen = 7.
• Committee must include 3 ladies and 4 gentlemen.
• Forbidden: committees that include both Mrs. X and Mr. Y together.

Concept / Approach:Use total-minus-forbidden counting.

Step-by-Step Solution:Unrestricted committees: C(8, 3) × C(7, 4) = 56 × 35 = 1960.Forbidden committees (contain both X and Y): choose the remaining 2 ladies from the other 7 → C(7, 2) = 21; choose the remaining 3 gentlemen from the other 6 → C(6, 3) = 20.Forbidden = 21 × 20 = 420.Valid committees = 1960 − 420 = 1540.

Verification / Alternative check:Partition by whether Y is included: when Y is included, exclude X; when Y is not included, X is free — this yields the same total.

Why Other Options Are Wrong:1960 ignores the restriction; 3240 exceeds the total; other values do not match the subtraction.

Common Pitfalls:Subtracting committees that contain X but not Y (which are allowed) by mistake.

Final Answer:1540