Desk posting puzzle and a Friday change that breaks the conditions
Five newly recruited assistants - P, Q, R, S and T - are to be posted at five verification desks numbered 1 to 5 in one row from left to right, one assistant per desk. The manager's note fixing the posting records only four conditions: T occupies desk 5; R sits exactly two desks to the right of S; Q sits somewhere to the left of R; and P occupies an even-numbered desk. A senior clerk who has not seen the full note tells the trainees that the order in which the conditions are written does not matter, and that a second, different posting plan also satisfies all of them. Separately, the audit team will visit every Friday, and the manager now wants desk 5 kept vacant on those days so that the team can work there, with T shifted to desk 2 for that day only and every other posting left unchanged.
1. Determine the full posting plan and show how each alternative is eliminated.
Read the worked answer
Step 1 - fix T. T is at desk 5.
Step 2 - place R and S. R is exactly two desks right of S, so the possible pairs are (S1, R3), (S2, R4) and (S3, R5). The third is impossible because desk 5 is T's. That leaves two branches.
Step 3 - test the branches against P's even desk.
- Branch (S2, R4): even desks are 2 and 4, but desk 2 is already S and desk 4 is R, so P has no even desk. Rejected.
- Branch (S1, R3): even desks 2 and 4 are still free, so P can take one of them.
Step 4 - place Q. With S1, R3 and T5, the free desks are 2 and 4. Q must be left of R (desk 3), which forces Q to desk 2, leaving desk 4 to P, which satisfies the even-desk condition.
Final plan: desk 1 S, desk 2 Q, desk 3 R, desk 4 P, desk 5 T. It is the only plan that satisfies all four conditions.
2. Is the senior clerk right that a second plan satisfies the conditions, and does the order of the conditions affect the answer?
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No second plan exists. Working through the branches above shows that once T is fixed at 5 and R must be two places right of S, the only branch that also allows an even desk for P is S1–R3, and the remaining two desks are then fully determined by Q's position relative to R. There is no scope for a second arrangement.
On ordering. The clerk is right only in a trivial sense: the four conditions are cumulative, so rewriting them in a different order cannot change which arrangements satisfy all of them - the solution set stays the same. But order is not irrelevant to the work: the efficient sequence is to fix T first (it removes one branch immediately), then expand the R–S pairs, then filter with P's even desk, and only then apply Q's left-of-R condition. A candidate who starts with Q wastes time, because Q's condition by itself eliminates nothing.
Practical point. A posting note with four conditions is not the same as a posting plan; the clerk should not treat the plan as verified until the branch-elimination has been done and recorded.
3. If desk 5 is left vacant on Fridays and T alone is moved to desk 2, does the plan still work? What should the manager be told?
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It fails, and not merely in one detail. With T moved to desk 2 and S1, R3, P4 unchanged, Q must sit somewhere left of R (desk 3), and the only desk left on that side is desk 2 - which T now occupies. Q also cannot take desk 5, since that is the desk being vacated. So the condition 'Q is left of R' cannot be satisfied.
More fundamentally, five assistants and five desks mean that vacating one desk leaves one assistant with no position in the row; the instruction is internally inconsistent, not just awkwardly worded.
What to tell the manager: on Fridays either (a) place the audit team outside the verification row, or (b) re-derive the whole plan with a spare sixth desk so that the two-places-apart condition can be met in the extended row, rather than shifting T by itself. Changing one posting in a linked arrangement almost always breaks a neighbour's condition, so any single-desk change should be re-tested against all four conditions before it is circulated.