Five Speeches and a Printer Deadline
Five student council candidates - A, B, C, D and E - will each give a three-minute speech, one per slot, in slots 1 to 5. The adviser sets these rules:
- A must speak immediately before B.
- C must speak before D.
- E cannot speak in slot 1 or slot 5.
- D cannot speak in slot 1. No two candidates share a slot. The program printer needs a final order today, but the adviser also asks whether the rules force a single possible order or leave room for choice. Two candidates have already asked to swap slots, and the adviser wants to know whether any swap is even permissible under the rules.
1. Do the rules determine one unique speaking order? Determine how many valid orders exist and list them.
Read the worked answer
The rules do not force a unique order. Treat A and B as a single block [AB], then test each possible block position.
- [AB] in slots 1–2: slots 3, 4, 5 hold C, D, E with C before D and E not in 5 → E must be 3 or 4. Gives A B E C D and A B C E D.
- [AB] in slots 2–3: slots 1, 4, 5 hold C, D, E. E cannot be 1 or 5, so E = 4. Then C and D fill 1 and 5; D cannot be 1, so C = 1, D = 5 → C A B E D.
- [AB] in slots 3–4: slots 1, 2, 5 hold C, D, E. E cannot be 1 or 5, so E = 2. Then C = 1, D = 5 → C E A B D.
- [AB] in slots 4–5: slots 1, 2, 3 hold C, D, E. E is 2 or 3.
- E = 2 → C and D in 1 and 3; D not 1, so C = 1, D = 3 → C E D A B.
- E = 3 → C and D in 1 and 2; D not 1, so C = 1, D = 2 → C D E A B.
Total: 6 valid orders. The printer needs an additional decision from the adviser, not just the rules.
2. If B speaks in slot 3, what is the full order? Show that no other arrangement survives.
Read the worked answer
B in slot 3 forces the [AB] block into slots 2–3, so A = 2, B = 3.
Remaining slots: 1, 4, 5 for C, D, E.
- E cannot be 1 (rule) or 5 (rule), so E must be 4. The only slot left for E in {1, 4, 5} is 4.
- Slots 1 and 5 remain for C and D, with C before D.
- The only way C precedes D across those two slots is C = 1, D = 5. This also satisfies D ≠ 1.
Full order: C A B E D.
This holds for every branch: E's restriction removes slot 1, its own rule removes slot 5, leaving a single position. Once E is placed, the C-before-D constraint plus the D ≠ 1 constraint pin the other two. So the answer is unique under this added condition.
3. Can C ever speak in slot 5? Justify your answer without listing all six orders.
Read the worked answer
No.
The rule states C must speak before D. If C were in slot 5, D would have to occupy a slot strictly after slot 5 - that is, slot 6 or higher. No such slot exists in a five-slot program.
So C in slot 5 leaves D with zero legal positions, and the schedule collapses. This is a general property worth noting: whenever a constraint says "X before Y," X can never take the last slot in the sequence, regardless of what the other rules say. The same reasoning shows D can never take slot 1, which is exactly why the adviser wrote that rule explicitly.
A useful check on the enumerated list: C appears in slots 1, 1, 1, 1, 1, 4 across the six valid orders - never in slot 5, confirming the deduction.