Six Counters, Six Clerks: Fixing the Posting Order
The district administration has set up six single-window counters, numbered 1 to 6 from the entrance to the back wall, and six clerks - Amar, Bindu, Chetan, Divya, Ekta and Farid - are to be posted one to a counter. The posting must satisfy these conditions:
- Farid must be at counter 6.
- Amar must be at the counter immediately to the left of Chetan.
- Divya must be at the counter immediately to the right of Bindu.
- Ekta must not be at a counter adjacent to Farid.
- Chetan must not be at counter 2.
- Divya must not be at a counter adjacent to Ekta.
- Bindu must not be at counter 2.
The assistant commissioner wants the plan finalised before the counters open, and has asked for the full order plus a verification of two claims made by a clerk.
1. Determine the complete posting order, showing how the alternatives are eliminated.
Read the worked answer
Farid is fixed at 6, so the other five occupy counters 1–5. Amar–Chetan and Bindu–Divya are two blocks of two adjacent counters, so together they use four of counters 1–5 and Ekta takes the remaining one.
The only ways to fit two disjoint pairs into 1–5 are: pairs at (1,2) and (3,4), leaving Ekta at 5; pairs at (1,2) and (4,5), leaving Ekta at 3; or pairs at (2,3) and (4,5), leaving Ekta at 1.
- Ekta at 5 fails, because Ekta would be adjacent to Farid at 6.
- Ekta at 3: if Amar–Chetan is at (1,2), then Chetan is at 2, which is barred. If instead Bindu–Divya is at (1,2), then Divya is at 2 and Ekta at 3, which is barred.
- So Ekta is at 1, and the pairs fill (2,3) and (4,5). Putting Bindu at 2 is barred, so Bindu–Divya must be (4,5) and Amar–Chetan (2,3).
Final order: Ekta 1, Amar 2, Chetan 3, Bindu 4, Divya 5, Farid 6.
2. A clerk claims that (i) exactly one clerk sits between Amar and Bindu, and (ii) Divya is posted at an end counter. Test both claims against the final order.
Read the worked answer
Using the order Ekta 1, Amar 2, Chetan 3, Bindu 4, Divya 5, Farid 6:
Claim (i). Amar is at 2 and Bindu at 4. The counters strictly between them hold only Chetan, at 3. So exactly one clerk sits between Amar and Bindu - the claim is true.
Claim (ii). The end counters are 1 and 6, occupied by Ekta and Farid. Divya is at 5, which is next to the end but not an end. The claim is false.
A useful check when a claim is about 'ends': in this arrangement neither of the two-block members (Amar, Chetan, Bindu, Divya) occupies an end, because both ends are forced to Ekta and Farid by the adjacency clause on Ekta.
3. If a seventh counter is added at the far end and Ekta is shifted there, the relative order of the other five staying unchanged, who is now two places to the right of Chetan, and who is adjacent to Farid?
Read the worked answer
Ekta moves from counter 1 to counter 7, and the other five close up while keeping their sequence: Amar 1, Chetan 2, Bindu 3, Divya 4, Farid 5, counter 6 vacant, Ekta 7.
Two places to the right of Chetan. Chetan is at 2; two places to the right is counter 4, which is Divya.
Adjacent to Farid. Farid is at 5, so the neighbouring counters are 4 and 6. Counter 4 holds Divya and counter 6 is vacant. So only Divya is adjacent to Farid.
Note that the original clue that Ekta must not be adjacent to Farid is still honoured: Ekta is at 7 and Farid at 5, with a vacant counter between them.