Six Desks, Three Fixed Rules
The Bundelkhand Gramin branch must roster six probationary officers - P, Q, R, S, T and U - at six adjacent desks, D1 to D6, numbered left to right, for a single audit day. Three rules are fixed and cannot be relaxed: R sits at D3, the desk that carries the cash-verification terminal; S sits immediately to the right of Q, because the pair share one key-dispensing tray; and T avoids both end desks, since the end desks face the public counter and T is still under observation. A written regional-office instruction adds that U must sit somewhere to the left of P. Each desk holds exactly one officer and each officer holds exactly one desk. An assistant has drafted a roster and claims the rules permit exactly one arrangement. The branch head doubts this and wants the count verified before printing, because a wrong roster would stall the day's cash verification and push the audit past its deadline.
1. How many valid rosters exist? Show the full deduction, not just the final number.
Read the worked answer
Step 1 - fix what is fixed. R occupies D3. The Q–S pair must be adjacent with S on the right, so candidate pairs are (D1,D2), (D2,D3), (D3,D4), (D4,D5), (D5,D6). The pairs (D2,D3) and (D3,D4) are impossible because D3 is already R's. Three legal blocks remain: Q1-S2, Q4-S5, Q5-S6.
Step 2 - fill each case. T cannot take D1 or D6; U must be left of P.
- Block Q1-S2: free desks D4, D5, D6. T can be D4 (then U5-P6) or D5 (then U4-P6). → 2 rosters.
- Block Q4-S5: free desks D1, D2, D6. T can only be D2, since D1 and D6 are end desks. Then U1-P6. → 1 roster.
- Block Q5-S6: free desks D1, D2, D4. T can be D2 (then U1-P4) or D4 (then U1-P2). → 2 rosters.
Total: 2 + 1 + 2 = 5 valid rosters. The assistant's claim of a unique arrangement is false, and the error is traceable to the two Q5-S6 branches that are easy to overlook.
2. The branch head now insists that T must occupy D4. How many rosters survive, and which ones?
Read the worked answer
Test the five rosters against the new condition T = D4:
- Q1-S2-R3-T4-U5-P6 → survives (ends are Q and P, so the end-desk rule holds).
- Q1-S2-R3-U4-T5-P6 → fails, T is at D5.
- U1-T2-R3-Q4-S5-P6 → fails, T is at D2.
- U1-T2-R3-P4-Q5-S6 → fails, T is at D2.
- U1-P2-R3-T4-Q5-S6 → survives.
Answer: 2 rosters.
The common trap is to keep the Q4-S5 block alive; in that block T is forced to D2 by the end-desk rule, so fixing T at D4 kills the branch outright rather than merely shifting a name. Note also that adding a condition removed three of five rosters - a single seat restriction can cut a puzzle's solution space far more sharply than a positional one.
3. An officer argues: "If U is placed at D1, the roster is fully determined." Is that correct? Justify with the surviving arrangements.
Read the worked answer
No. Filter the five rosters for U at D1:
- Q1-S2-R3-T4-U5-P6 → U is at D5, excluded.
- Q1-S2-R3-U4-T5-P6 → U is at D4, excluded.
- U1-T2-R3-Q4-S5-P6 → survives.
- U1-T2-R3-P4-Q5-S6 → survives.
- U1-P2-R3-T4-Q5-S6 → survives.
Three rosters survive, so U at D1 does not determine the arrangement. The reason is structural: U is already pushed to D1 in every case except the Q1-S2 block, so the condition eliminates one block but leaves the choice between T at D2 and T at D4, and between Q at D4 and Q at D5, untouched. A useful check for data-sufficiency questions of this type: ask whether the added condition discriminates between the branches of the solution tree, not merely between individual rosters. Here it eliminates one branch out of three and leaves two internal choices open.