2016 AMC 10A Problem 13

Track the sum of the seat numbers occupied by the five people. Before and after, the five of them fill all five seats, so the total is $1+2+3+4+5 = 15$ both times.

Add up everyone's change in seat number:
- Bea: $+2$.
- Ceci: $-1$.
- Dee and Edie swapped, so their changes cancel: $0$.
- Ada: unknown, call it $a$.

Since the total does not change, $2 - 1 + 0 + a = 0$, so $a = -1$: Ada's new seat is one to the left of her old seat.

Ada's new seat is an end seat, $1$ or $5$. If it were $5$ she would have come from seat $6$, which does not exist. So she now sits in seat $1$ and originally sat in seat $2$.

A consistent arrangement: originally Bea $1$, Ada $2$, Ceci $3$, Dee $4$, Edie $5$; afterwards Ceci $2$, Bea $3$, Edie $4$, Dee $5$, with seat $1$ open for Ada.

The answer is $\boxed{\textbf{(B)}\ 2}$.

When people permute among a fixed set of seats, the sum of occupied seat numbers is invariant, so the individual displacements must sum to zero. That single equation pins down the one unknown displacement without any trial and error. Look for such conserved totals whenever a puzzle describes several moves and asks about one of them.

Trial: suppose Ada ends in seat $5$; then Bea (moved right by $2$) could not have started in seats $4$ or $5$, Ceci (moved left) could not have started in $1$, and no assignment works. Suppose Ada ends in seat $1$; the arrangement above works, and her start was seat $2$.

• Assuming Ada must return to seat 5 (the right end) or forgetting that Dee and Edie's swap contributes zero net movement.
• Reading "left" and "right" as the audience's reversed view; the problem states directions are from the sitters' perspective, matching the seat numbering.