Six-Floor Bank Office Allocation
A cooperative bank is assigning six officers-Asha, Bala, Chitra, Deepak, Esha and Farhan-to six floors numbered 1 to 6, one officer per floor. Floor 1 is the ground floor and floor 6 is the top floor. The branch has fixed four conditions. Chitra must sit on floor 4. Asha must be immediately above Bala. Esha must be exactly two floors above Farhan. Deepak cannot be on floor 1 or floor 6. No officer may share a floor. The administrative officer has to publish the final floor chart before a fire drill, so the placement must satisfy every condition at once. A staff member has proposed that Farhan be placed on floor 3, but the proposal has not yet been checked. The branch also wants to know whether the chart would change if the restriction on Deepak were removed.
1. Determine the complete floor arrangement, showing how each constraint eliminates alternatives.
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Start with Chitra = floor 4.
- Asha is immediately above Bala, so possible pairs are (Bala, Asha) = (1,2), (2,3), (3,4), (4,5), (5,6). Because Chitra occupies floor 4, pairs (3,4) and (4,5) are impossible. Remaining: (1,2), (2,3), (5,6).
- Esha is exactly two floors above Farhan, so possible pairs are (Farhan, Esha) = (1,3), (2,4), (3,5), (4,6). Chitra on 4 rules out (2,4) and (4,6). Remaining: (1,3) and (3,5).
Test the Asha-Bala pairs:
- If Bala = 1 and Asha = 2, then Esha-Farhan must be (3,5), leaving Deepak on 6. This violates “Deepak cannot be on floor 6.”
- If Bala = 2 and Asha = 3, then floor 3 is already occupied by Asha, so Farhan cannot be on 3; and Esha cannot be on 3. No Esha-Farhan pair works.
- Therefore Bala = 5 and Asha = 6. The Esha-Farhan pair cannot be (3,5) because Bala is on 5, so Farhan = 1 and Esha = 3. The only remaining floor for Deepak is 2, which is allowed.
Final chart: floor 1 Farhan; floor 2 Deepak; floor 3 Esha; floor 4 Chitra; floor 5 Bala; floor 6 Asha.
2. Can Farhan be placed on floor 3? Explain which constraint fails.
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No. If Farhan is on floor 3, then Esha must be exactly two floors above him, so Esha = floor 5. Chitra is fixed on floor 4. The remaining floors for Asha, Bala and Deepak are 1, 2 and 6. For Asha to be immediately above Bala, the only available pair among those floors is Bala = 1 and Asha = 2, because Bala = 2 would require Asha on 3, which is occupied by Farhan, and Bala = 5 would require Asha on 6 but floor 5 is occupied by Esha. With Bala = 1 and Asha = 2, Deepak is forced to floor 6, which violates the rule that Deepak cannot be on floor 6. So Farhan cannot be on floor 3.
3. If the Deepak restriction is removed, how many valid arrangements exist? Identify the extra arrangement.
Read the worked answer
Removing the Deepak restriction affects only the first branch that was earlier rejected. With the other conditions unchanged:
- Bala = 1, Asha = 2, Farhan = 3, Chitra = 4, Esha = 5, Deepak = 6 is now valid.
- The original arrangement Bala = 5, Asha = 6, Farhan = 1, Deepak = 2, Esha = 3, Chitra = 4 remains valid.
- The middle Asha-Bala pair (Bala = 2, Asha = 3) is still impossible because floor 3 is needed by Asha and Farhan cannot then be placed.
So there are exactly two valid arrangements. The extra one is: floor 1 Bala; floor 2 Asha; floor 3 Farhan; floor 4 Chitra; floor 5 Esha; floor 6 Deepak.