The Branch Audit Roster That Would Not Settle
Six probationary assistants - Anil, Bina, Chetan, Divya, Esha and Farhan - must each audit exactly one of six branches P, Q, R, S, T and U, one branch per day from Monday to Saturday. The branch head's note records: (i) Bina audits Q on Wednesday. (ii) Chetan audits R, exactly two days after Farhan's audit. (iii) Divya audits S. (iv) Esha audits T, but neither on Monday nor on Saturday. (v) Anil, who does not audit P, is scheduled before Divya. (vi) No two assistants audit on the same day. A clerk claims the roster is fully determined by these six statements and prepares a single day-wise plan. Before approving it, the branch head asks you to test the claim and to accommodate a late request from the regional office: the S-branch audit should be held as early in the week as possible.
1. Is the clerk's claim correct? Work through the possibilities and state how many valid rosters exist.
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Bina is fixed on day 3 (Wednesday, Q). Chetan audits R and his day is Farhan's day + 2. Farhan can only be day 1, 2, 4 or 5, because day 3 is taken; Chetan would then be day 3, 4, 6 or 7. Day 3 is occupied and day 7 does not exist, so Farhan is day 2 or day 4.
Branches Q, R, S and T are taken, so Anil and Farhan must take P and U. Since Anil does not audit P, Anil = U and Farhan = P in every valid roster.
If Farhan = day 2, Chetan = day 4. Remaining days for Anil, Divya and Esha are 1, 5 and 6. Esha cannot be day 1 or day 6, so Esha = day 5. Anil before Divya then forces Anil = day 1 and Divya = day 6. That gives one roster.
If Farhan = day 4, Chetan = day 6. Remaining days are 1, 2 and 5. Esha can be day 2 or day 5. Anil before Divya gives two rosters: (Anil 1, Divya 2, Esha 5) and (Anil 1, Divya 5, Esha 2).
So there are three valid rosters, not one. The clerk has mistaken a consistent partial solution for a unique solution.
2. The regional office wants Divya's audit as early as possible. Which roster satisfies this, and on what day does Divya audit?
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Divya cannot audit on Monday, because Anil must be scheduled before Divya and no day precedes Monday. The earliest possible day for Divya is therefore Tuesday (day 2).
The roster that achieves this is: Monday - Anil (U); Tuesday - Divya (S); Wednesday - Bina (Q); Thursday - Farhan (P); Friday - Esha (T); Saturday - Chetan (R).
Check each condition: Bina is on Wednesday at Q; Chetan is exactly two days after Farhan (Thursday to Saturday); Divya is at S; Esha is at T on Friday, not Monday or Saturday; Anil is before Divya and does not audit P; all six days are distinct. Every condition holds, so Tuesday is achievable and is the minimum.
3. A clerk suggests fixing Esha's audit on Thursday to make the roster unique. Test the suggestion.
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The suggestion fails. Thursday is day 4. In the roster where Farhan is day 2, Chetan already occupies day 4. In the other two rosters, Farhan himself occupies day 4. Thus every valid roster uses day 4 for either Chetan or Farhan, and the one-branch-per-day rule prevents Esha from being placed there as well. Adding 'Esha on Thursday' makes the puzzle unsatisfiable rather than unique.
A condition that would genuinely make the roster unique is 'Divya audits on Saturday' or, equivalently, 'Farhan audits on Tuesday'. Either condition selects only the first roster, in which Anil = Monday, Divya = Tuesday, Bina = Wednesday, Farhan = Thursday, Esha = Friday and Chetan = Saturday.