The Practicum Allocation Chart That Must Be Right
Kavita coordinates the school internship at a B.Ed college that has a four-week placement tie-up with a nearby school. Five student-teachers - Asha, Bhavna, Chetan, Divya and Eshaan - must each be assigned to exactly one cooperating teacher, and the school has allotted exactly one class each from grades 5, 6, 7, 8 and 9. The school's scheduling sheet records only three rules, and all of them hold: (i) Divya teaches grade 6; (ii) Bhavna's grade number is exactly one higher than Chetan's; (iii) Asha is placed neither in grade 5 nor in grade 9. The principal wants the final allocation chart by tomorrow morning. A senior colleague has already circulated a draft that places Eshaan in grade 7 and Divya in grade 6, and has told Kavita that the rest 'can be adjusted later'. The school will print the chart on its notice board, so a wrong allocation will be visible to parents, and the cooperating teachers will build their unit plans around it.
1. Produce the complete allocation and show how the three rules force it. State whether the solution is unique.
Read the worked answer
Step 1 - narrow the Bhavna/Chetan pair. Rule (ii) allows only (5,6), (6,7), (7,8) or (8,9). Rule (i) puts Divya in grade 6, and each grade is used once, so (5,6) and (6,7) are both impossible. Pair is either (7,8) or (8,9).
Step 2 - test pair (7,8). Bhavna = 7, Chetan = 8. Grades 5 and 9 remain for Asha and Eshaan. Rule (iii) forbids Asha in 5 and in 9, so Asha has no grade. Dead end.
Step 3 - test pair (8,9). Bhavna = 8, Chetan = 9. Grades 5 and 7 remain. Asha cannot take 5, so Asha = 7 and Eshaan = 5.
Result: Asha 7, Bhavna 8, Chetan 9, Divya 6, Eshaan 5.
Uniqueness: the two pair-cases exhaust rule (ii), and exactly one survives rules (i) and (iii), so the solution is unique - no other chart satisfies all three rules. The colleague's draft placing Eshaan in grade 7 is therefore wrong: grade 7 is forced to Asha.
2. Rule (iii) is withdrawn. How many allocations now satisfy the remaining rules? Show the counting rather than guessing.
Read the worked answer
With only rules (i) and (ii), the Bhavna/Chetan pair is again restricted to (7,8) or (8,9), because grade 6 is taken by Divya.
- Pair (7,8): grades 5 and 9 remain for Asha and Eshaan. Two orders: Asha 5 / Eshaan 9, or Asha 9 / Eshaan 5.
- Pair (8,9): grades 5 and 7 remain for Asha and Eshaan. Two orders: Asha 5 / Eshaan 7, or Asha 7 / Eshaan 5.
Total = 2 + 2 = 4 valid allocations.
Note what this tells Kavita: rule (iii) was doing almost all the work of pinning the chart down. Removing it multiplies the solutions fourfold, which is why the chart cannot be finalised from the colleague's partial draft - 'adjusting the rest later' would mean choosing among four different charts, and three of them conflict with a rule the school has already published.
3. The principal proposes a fourth rule: 'Eshaan must be placed in an even-numbered grade.' Evaluate the proposal and advise Kavita on what to say.
Read the worked answer
Test it against the constrained system, not in isolation.
- From the earlier solution, rules (i)-(iii) already force Eshaan into grade 5.
- 5 is odd, so rule (iv) contradicts the existing three rules.
- Check whether dropping rule (iii) rescues it: with only (i) and (ii), the surviving pairs are (7,8) and (8,9). In pair (7,8) the spare grades are 5 and 9 - both odd. In pair (8,9) they are 5 and 7 - both odd. So rule (iv) is unsatisfiable even without rule (iii).
Conclusion: adding rule (iv) leaves zero valid allocations; it is not a redundant rule but an impossible one.
Advice to Kavita: say that a constraint set must be checked for joint satisfiability before it is announced, and that the draft on the notice board cannot be 'fixed later' by adding wishes. Concretely, propose either dropping rule (iv) or renegotiating rule (ii) - for example, allowing Bhavna's grade to be two higher than Chetan's - and then re-solving. Presenting the failed case split is stronger than saying the idea 'seems wrong'.