Scheduling Six Trainees Across Three Sessions
Six trainees - A, B, C, D, E and F - must be assigned to exactly three induction sessions held on the same day: Session 1 at 9:00 a.m., Session 2 at 11:00 a.m. and Session 3 at 2:00 p.m. Each session takes exactly two trainees, and each trainee attends exactly one session. The coordinator's rules are: (i) A and B must not attend the same session; (ii) C must attend the session immediately before the one D attends; (iii) E must attend Session 3; (iv) F must attend Session 2; (v) A must attend Session 1. The coordinator has produced a draft in which C and F are placed in Session 2, D and E are placed in Session 3, and A and B are left in Session 1. She states that the draft satisfies every rule and asks for it to be published.
1. Check the draft against all five rules and produce the correct assignment, showing the order in which the deductions follow.
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The draft fails. C in Session 2 and D in Session 3 does satisfy rule (ii), E in Session 3 satisfies (iii), F in Session 2 satisfies (iv), and A in Session 1 satisfies (v). But the draft forces B into Session 1 with A, which breaks rule (i). So the draft is invalid.
Correct deduction: rule (ii) allows only two placements - C in Session 1 with D in Session 2, or C in Session 2 with D in Session 3. Suppose the second: Session 2 would hold C and F (rule iv), leaving D and E for Session 3 (rule iii). Session 1 is then A and B together - impossible under rule (i). So C must be in Session 1 and D in Session 2.
Session 2 is now D and F. Session 1 is C plus one of A or B, but rule (v) puts A in Session 1, so Session 1 is A and C. The remaining trainee, B, joins E in Session 3.
Final assignment: Session 1 - A, C; Session 2 - D, F; Session 3 - B, E. Every rule is satisfied.
2. The coordinator now wants F in Session 3 and E in Session 2, with A still in Session 1 and all other rules unchanged. Does a valid assignment exist? Justify your answer.
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Yes, and it is unique.
With E fixed in Session 2 and F in Session 3, test the two placements allowed by rule (ii).
Case 1 - C in Session 1, D in Session 2. Session 1 already contains A (rule v), so it becomes A and C. Session 2 is D and E. Session 3 is F plus the one remaining trainee, B. Check rule (i): A and B are in Sessions 1 and 3, so they are separated. Valid.
Case 2 - C in Session 2, D in Session 3. Session 2 becomes C and E; Session 3 becomes D and F. That leaves A and B together in Session 1, which breaks rule (i). Invalid.
So exactly one assignment works: Session 1 - A, C; Session 2 - D, E; Session 3 - B, F.
Notice what changed from the original answer: only the identities in Sessions 2 and 3 moved. C and D stayed adjacent, and A stayed in Session 1, which is why the schedule survives the swap.
3. If rule (i) were withdrawn but rule (v) kept, how many assignments would satisfy the remaining rules? List them.
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Two assignments.
Without rule (i), A and B may share a session, but rule (v) still pins A to Session 1, and rules (iii) and (iv) still pin E to Session 3 and F to Session 2.
Case 1 - C in Session 1, D in Session 2. Session 1 holds A and C, Session 2 holds D and F, Session 3 holds B and E. This is the same schedule as in the original problem.
Case 2 - C in Session 2, D in Session 3. Session 2 becomes C and F, Session 3 becomes D and E, and the two remaining trainees, A and B, must take Session 1. Rule (v) is satisfied because A is in Session 1. Under the old rules this case was excluded by rule (i), since A and B would be together; with rule (i) withdrawn, it is legal.
No other cases exist, because rule (ii) permits only those two placements of C and D, and the rest of the schedule is then forced by rules (iii), (iv) and (v). The answer is therefore exactly two, not one: withdrawing a constraint loosens the schedule without creating new placement options for C and D.