For which of the following children is it possible to determine how many times they received the Buck?

The displacements of rounds $3,6,7,8,9$ are read straight from their given pass types: $+2,-2,-2,-2,+1$.

Bina is fixed at seat $0$.

Round $5$ delivers to Aarav; call Aarav's seat $x$.

Round $6$ has $\Delta=-2$, so Eshan sits at $x-2$.

Round $7$ has $\Delta=-2$, so Gaurav sits at $x-4$.

Continuing the chain:

Round $8$, $\Delta=-2$: Chirag at $x-6 \equiv x+1$.

Round $9$, $\Delta=+1$: Diya at $x+2$.

Diya is also the round-$2$ receiver, so round $2$ lands on $x+2$.

Round $3$ has $\Delta=+2$ from Diya, so Farhan sits at $x+4$.

Round $10$ runs from Diya $(x+2)$ to Chirag $(x+1)$, a displacement of $-1$, so its pass type is immediately to the right.

The six children Aarav, Chirag, Diya, Gaurav, Farhan, Eshan occupy seats $x,\,x+1,\,x+2,\,x+3,\,x+4,\,x+5$ (since $x-4\equiv x+3$ and $x-2\equiv x+5$) — six consecutive seats.

The only remaining seat is $x+6\equiv x-1$, which must be Bina.

Bina $=0$ gives $x-1\equiv 0$, so $x=1$.

Seating (seat : child):

$0$ Bina, $1$ Aarav, $2$ Chirag, $3$ Diya, $4$ Gaurav, $5$ Farhan, $6$ Eshan.

The fixed-round seats are filled in accordingly.

Rounds $1$ and $2$ carry the buck from Bina $(0)$ to Diya $(3)$ through seat$_1$.

seat$_1$ must be within two of $0$: $\{1,2,5,6\}$, and within two of $3$: $\{1,2,4,5\}$.

The overlap is $\{1,2,5\}$, so the round-$1$ receiver is Aarav $(1)$, Chirag $(2)$ or Farhan $(5)$.

The matching displacement pairs $(\Delta_1,\Delta_2)$ are $(+1,+2),\,(+2,+1),\,(-2,-2)$ — none is $-1$.

Rounds $4$ and $5$ carry the buck from Farhan $(5)$ to Aarav $(1)$ through seat$_4$.

seat$_4$ must be within two of $5$: $\{6,0,4,3\}$, and within two of $1$: $\{0,2,6,3\}$.

The overlap is $\{0,3,6\}$, so the round-$4$ receiver is Bina $(0)$, Diya $(3)$ or Eshan $(6)$.

The displacement pairs $(\Delta_4,\Delta_5)$ are $(+2,+1),\,(-2,-2),\,(+1,+2)$ — again none is $-1$.

Counting displacements:

The value $-1$ (immediately to the right) occurs only in round $10$; neither free window can produce it.

So the immediately-to-the-right total is exactly $1$ and is forced.

$+1$ (round $9$ plus windows), $+2$ (round $3$ plus windows) and $-2$ (rounds $6,7,8$ plus windows) all shift with the window choices, so their totals are not forced.

Counting receivers:

The fixed rounds give Bina $0$, Aarav $1$, Chirag $2$, Diya $2$, Gaurav $1$, Farhan $1$, Eshan $1$.

Round $1$ adds one to whichever of Aarav/Chirag/Farhan it hits; round $4$ adds one to whichever of Bina/Diya/Eshan it hits.

Gaurav lies in neither window, so his total stays at exactly $1$ — the only forced receive total. Every other child's total can move by $1$.

Final seating: $0$ Bina, $1$ Aarav, $2$ Chirag, $3$ Diya, $4$ Gaurav, $5$ Farhan, $6$ Eshan. The seats are fully fixed; only the round-$1$ and round-$4$ receivers (with their linked displacements) stay interchangeable.

The fixed rounds give each child a base count, with Gaurav (seat $4$) received exactly once (round $7$). Seat $4$ falls outside both ambiguous windows: round $1$ can add one only to Aarav, Chirag or Farhan, and round $4$ can add one only to Bina, Diya or Eshan. So every child except Gaurav can gain an extra reception, while Gaurav's total is locked at $1$. Only Gaurav's count is determinable — option 2.