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Seven people A, B, C, D, E, F, and G have meetings on different days of a week starting from Monday to Sunday. Only two people have meetings between A and C. F has a meeting on Wednesday. B has a meeting immediately after C. D has a meeting before A. How many people have meetings after G?
0
1
2
3
0
Fix the anchor point (F at Wednesday) and map the surrounding blocks (C, B) first to narrow down the limited possibilities for D and A.
Fix the anchor point (F at Wednesday) and map the surrounding blocks (C, B) first to narrow down the limited possibilities for D and A.
Establish the base grid
Days: Mon, Tue, Wed, Thu, Fri, Sat, Sun. We are given F is Wednesday. The constraint 'Only two people between A and C' with 'B immediately after C' implies the blocks (C, B) or (A, _, _, C, B) or (C, B, _, _, A).
Applying constraints
If C is Mon, B is Tue, then A must be Thu (two people between A and C). This leaves E, D, G for Fri, Sat, Sun. D is before A, but if A is Thu, D must be earlier. D must be before A, and F is Wed. Placing D on Mon/Tue/Thu creates a conflict. Let's test the configuration: Mon: D, Tue: A, Wed: F, Thu: E, Fri: C, Sat: B, Sun: G. Here, two people are between A and C (F, E), and B follows C.
Final verification
Final sequence: Mon: D, Tue: A, Wed: F, Thu: E, Fri: C, Sat: B, Sun: G. All conditions satisfied: 1. Two between A and C (F, E). 2. F is Wed. 3. B immediately after C (Fri/Sat). 4. D before A. Result: G is on Sunday, so 0 people have meetings after G.
B is incorrect as only 0 people follow G. C is incorrect as only 0 people follow G. D is incorrect as only 0 people follow G.
A is correct because, based on the logical sequence D-A-F-E-C-B-G, G occupies the final day (Sunday), leaving no one after G.
In scheduling puzzles, always look for the most constrained variable first. If two people have a fixed gap between them, try placing the block in all valid positions immediately.