The Annual Function Timetable: Six Performers, Six Days
The annual function committee of a school must allot six solo performances - Aarav's dance, Bhavna's song, Chitra's skit, Dev's magic, Esha's recitation and Farhan's instrumental - to six consecutive days, Monday through Saturday, exactly one performance per day. The committee has already accepted four conditions. Chitra's skit must be on Thursday, because the auditorium's lighting rig is available only that evening. Dev must perform exactly two days after Aarav. Esha must perform on one of the three days before Chitra. Farhan cannot perform on Monday or Saturday, because he arrives late and leaves early for a family wedding. A fifth condition was minuted as "no two consecutive days may feature the same performer", which the secretary now concedes is meaningless. The principal wants an unambiguous timetable and has asked for exactly one further condition - no more - that removes the remaining ambiguity.
1. Using only the four substantive conditions, whose performance must fall on Saturday? Prove that your answer is forced rather than merely possible.
Read the worked answer
Number the days 1 (Mon) to 6 (Sat). Chitra = day 4.
Placing Aarav and Dev. Dev = Aarav + 2. Since Chitra occupies day 4, neither Aarav nor Dev can be on day 4, and Aarav cannot be on day 5 or 6 (then Dev would fall outside the week). So Aarav ∈ {1, 3}.
Case A - Aarav = 1, Dev = 3. Esha must be before day 4 and days 1 and 3 are taken, so Esha = 2. Days 5 and 6 remain for Bhavna and Farhan. Farhan cannot be on day 6, so Farhan = 5 and Bhavna = 6.
Case B - Aarav = 3, Dev = 5. Remaining days are 1, 2, 6 for Esha, Bhavna and Farhan. Farhan cannot be on 1 or 6, so Farhan = 2. Esha must then be = 1 (before 4), leaving Bhavna = 6.
Conclusion: only two timetables are possible, and in both of them Bhavna performs on Saturday. Farhan is the only performer who is not pinned to a single day - he lands on Friday in Case A and Tuesday in Case B, which is why "Farhan performs last" is the attractive wrong answer.
2. If the principal is told that Dev will perform on Friday, what is the complete day-by-day order?
Read the worked answer
Dev on Friday means Dev = day 5, so Aarav = day 3 (Dev is exactly two days after Aarav). Chitra remains on day 4 (Thursday).
That fixes Case B from the earlier analysis. The remaining slots are days 1, 2 and 6 for Esha, Bhavna and Farhan.
- Farhan cannot be on day 1 or day 6, so Farhan = day 2 (Tuesday).
- Esha must be before Chitra (day 4), and day 3 is taken, so Esha = day 1 (Monday).
- Bhavna = day 6 (Saturday).
Full order: Monday Esha (recitation), Tuesday Farhan (instrumental), Wednesday Aarav (dance), Thursday Chitra (skit), Friday Dev (magic), Saturday Bhavna (song).
A quick consistency check: each performer appears once, the skit is on Thursday, Dev is two days after Aarav, Esha is before Chitra, and Farhan avoids Monday and Saturday. All four conditions hold, and no alternative assignment survives once Dev is fixed.
3. Propose exactly one additional condition that would make the timetable unique, and demonstrate that it does the job.
Read the worked answer
The two surviving timetables differ in only one respect: whether Aarav or Esha occupies Monday.
- Case A: Mon Aarav, Tue Esha, Wed Dev, Thu Chitra, Fri Farhan, Sat Bhavna.
- Case B: Mon Esha, Tue Farhan, Wed Aarav, Thu Chitra, Fri Dev, Sat Bhavna.
So any single condition that separates Aarav's and Esha's positions will work. Several are equally valid:
- "Aarav performs before Esha." True in Case A (Aarav day 1, Esha day 2); false in Case B (Esha day 1). Uniquely selects Case A.
- "Dev performs on Wednesday." Case A yes, Case B no.
- "Esha performs on Monday." Uniquely selects Case B.
- "Farhan performs on Tuesday." Uniquely selects Case B.
A condition such as "Bhavna performs on Saturday" would be useless, because it is already true in both cases - it adds nothing. The test of a good extra condition is not that it is true, but that it is true in exactly one of the two surviving arrangements.