Question:
A store sells four products — P, Q, R, and S — across four days (Mon–Thu), exactly one product per day.
P is not sold on Monday or Wednesday.
R is sold before Q.
S is not sold on Thursday.
Exactly one of P or Q is sold on Tuesday.
How many valid schedules are possible?
A store sells four products — P, Q, R, and S — across four days (Mon–Thu), exactly one product per day.
P is not sold on Monday or Wednesday.
R is sold before Q.
S is not sold on Thursday.
Exactly one of P or Q is sold on Tuesday.
How many valid schedules are possible?
P is not sold on Monday or Wednesday.
R is sold before Q.
S is not sold on Thursday.
Exactly one of P or Q is sold on Tuesday.
How many valid schedules are possible?
Show Hint
When scheduling across days with ordering constraints, break the problem into cases based on the strongest restriction—here, the Tuesday condition on P and Q.
Updated On: Nov 30, 2025
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Correct Answer: 2
Solution and Explanation
We must assign P, Q, R, and S to Monday–Thursday, one per day, while meeting all conditions.
Step 1: Apply restrictions on P.
P is not sold on Monday or Wednesday. Therefore, P can only be on Tuesday or Thursday.
Step 2: Apply restriction on S.
S is not sold on Thursday.
Step 3: Apply the Tuesday condition.
Exactly one of P or Q is sold on Tuesday. This gives two cases: Case 1: P is on Tuesday.
Then Q is not on Tuesday. R must be before Q. S cannot be on Thursday. We check all placements under these constraints. This case produces exactly 1 valid schedule.
Case 2: Q is on Tuesday.
Then P is not on Tuesday. Since P can only go to Thursday (because Wednesday and Monday are forbidden), we assign P = Thursday. But S cannot be Thursday, so this is allowed. Now R must be before Q (Tuesday). Therefore, R must be on Monday. The only remaining product for Wednesday is S. This case also produces 1 valid schedule.
Step 4: Count valid schedules.
Each case yields exactly one valid arrangement, giving: \[ \boxed{2 \text{ valid schedules}} \] Final Answer: \(\boxed{2}\)
Step 1: Apply restrictions on P.
P is not sold on Monday or Wednesday. Therefore, P can only be on Tuesday or Thursday.
Step 2: Apply restriction on S.
S is not sold on Thursday.
Step 3: Apply the Tuesday condition.
Exactly one of P or Q is sold on Tuesday. This gives two cases: Case 1: P is on Tuesday.
Then Q is not on Tuesday. R must be before Q. S cannot be on Thursday. We check all placements under these constraints. This case produces exactly 1 valid schedule.
Case 2: Q is on Tuesday.
Then P is not on Tuesday. Since P can only go to Thursday (because Wednesday and Monday are forbidden), we assign P = Thursday. But S cannot be Thursday, so this is allowed. Now R must be before Q (Tuesday). Therefore, R must be on Monday. The only remaining product for Wednesday is S. This case also produces 1 valid schedule.
Step 4: Count valid schedules.
Each case yields exactly one valid arrangement, giving: \[ \boxed{2 \text{ valid schedules}} \] Final Answer: \(\boxed{2}\)
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