Five Children in a Row: When Two Answers Fit Every Clue
In a B.Ed. practice class, a teacher writes this puzzle on the board: five children-Asha, Bina, Chetan, Deepa and Esha-sit in one row facing north. Bina is not at either end. Chetan sits immediately to the right of Deepa. Asha is somewhere to the left of Bina. Esha is not adjacent to Bina. Three groups report only their final order. Group 1: Asha, Bina, Deepa, Chetan, Esha. Group 2: Esha, Asha, Bina, Deepa, Chetan. Group 3: Deepa, Chetan, Asha, Bina, Esha. With 12 minutes left, all three groups insist they are right. The teacher wants to add one clue that leaves exactly one group's order valid, without simply announcing the answer, because the learning goal is 'test every condition, not just the first one that fits'. She also hears a child in Group 2 say, 'Chetan must be at an end, because that is how our answer looks.'
1. Test each group's order against every clue. Which group or groups survive, and what does that tell the teacher about the puzzle as written?
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Test clue by clue, not by overall impression.
Group 1 = A, B, D, C, E: B is in position 2, so not at an end. C is immediately right of D. A is left of B. E is in position 5 and B is in position 2, so E is not adjacent to B. All four clues hold.
Group 2 = E, A, B, D, C: B is in position 3. C is right of D. A is left of B. E is position 1 and B is position 3, so E is not adjacent. All four clues hold.
Group 3 = D, C, A, B, E: B is position 4. C is right of D. A is left of B. But E is position 5 and B is position 4, so E is adjacent to B. This violates the last clue.
Judgement: Groups 1 and 2 both satisfy the stated clues, so the puzzle as written has at least two solutions. Group 3 is invalid. The teacher cannot declare a unique answer without an extra clue. This is the difference between 'my order fits' and 'my order is the only order that can fit'.
2. Propose one clue the teacher can add that leaves exactly one group valid, and explain why a plausible-looking alternative fails.
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One reliable clue is: 'Asha is not at either end.' Check it against the survivors. Group 1 has Asha in position 1, an end, so it is eliminated. Group 2 has Asha in position 2, so it survives. Group 3 already fails the adjacency clue. The added clue also removes the alternative valid order A, B, D, C, E, leaving E, A, B, D, C as the only order satisfying all five clues. Exactly one group remains.
A plausible-looking alternative fails: 'Bina is immediately right of Asha.' Group 1 has A then B, and Group 2 also has A then B, so both survive. Another weak choice is 'Esha is at an end': Group 1 has E at position 5 and Group 2 has E at position 1, so neither is eliminated. The design principle is that a discriminating clue must split the surviving candidates. A clue that is merely true of one group's answer does not discriminate.
3. A child says, 'Chetan must be at an end, because that is how our answer looks.' Is this a valid deduction from the four original clues? Name the reasoning error and suggest a teaching response.
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It is not a valid deduction. Group 1 is a counterexample: its order A, B, D, C, E satisfies all four original clues and has Chetan in position 4, not at an end. The child has taken a feature of their own answer and converted it into a necessity. This is a reverse inference, often described as affirming the consequent: from 'our answer has C at an end' the child concludes 'C must be at an end', although the clues never force that.
The conclusion may later become true if the teacher adds a clue such as 'Chetan is at an end', but the child's route to it is still unjustified. The teaching response is to ask: 'Which clue forces Chetan to an end?' and then ask the child to build a valid order with Chetan not at an end. That separates 'true in the answer' from 'forced by the clues', which is exactly the learning goal of the lesson.