2012 AMC 10B Problem 25

Set-up. Number the columns of right-pointing arrows from left to right. They hold $2, 2, 4, 4, 4, 2, 2$ arrows. Columns $2$, $4$ and $6$ also have one left-pointing arrow on the middle line.

1. Why stages help. Look at the far end of any arrow in columns $1$, $3$, $5$ or $7$. Every segment leaving it leads right, and the backward arrows only reach points between these columns. So once the bug passes through one of these arrows, it can never get back to anything on the left. That means every path uses exactly one arrow from each of columns $1$, $3$, $5$, $7$, and what the bug does after a checkpoint arrow doesn't depend on how it got there. We can count one stage at a time and multiply.
2. Column 1. There is $1$ way to reach each of its two arrows (straight from $A$).
3. Column 1 to column 3. Start at the end of the top column-1 arrow. The bug can:
- cross the top column-2 arrow and go on to either top-half column-3 arrow;
- slip down through the middle point to the bottom half, cross the bottom column-2 arrow, and climb back up through the next middle point to a top-half column-3 arrow;
- cross the top column-2 arrow, come down through the middle point, and go to a bottom-half column-3 arrow;
- slip down to the bottom half and cross the bottom column-2 arrow, reaching a bottom-half column-3 arrow directly;
- cross the top column-2 arrow, come down to the middle, take the backward arrow, drop to the bottom half and cross the bottom column-2 arrow.

Sorting these out, the top column-1 arrow leads to each top-half column-3 arrow in $2$ ways and to each bottom-half column-3 arrow in $3$ ways. The bottom column-1 arrow is the mirror image. So each column-3 arrow is reached in $2 + 3 = 5$ ways.
4. Column 3 to column 5. Listing routes the same way through the wider middle section: from a column-3 arrow there are $4$ routes to each column-5 arrow in the same half and $8$ routes to each one in the other half. Each column-5 arrow is fed by two same-half and two other-half column-3 arrows, so it is reached in
$$5 \times (4 + 4 + 8 + 8) = 5 \times 24 = 120 \text{ ways.}$$
5. Column 5 to column 7. From a column-5 arrow there are $2$ routes to the column-7 arrow in its half and $3$ to the other one. Each column-7 arrow is fed by four column-5 arrows, two from each half:
$$120 \times (2 + 2 + 3 + 3) = 120 \times 10 = 1200 \text{ ways.}$$
6. Finish. From each column-7 arrow there is exactly one way to $B$. With two column-7 arrows:
$$2 \times 1200 = 2400.$$

The answer is $\boxed{\textbf{(E)}\ 2400}$.

One-way arrows act like turnstiles: once the bug goes through certain ones, everything behind it is out of reach and no longer matters. That lets a huge count break into a product of small counts, one per stage. The up-down symmetry of the picture saves more work, since every arrow in the same half of a column gets the same count.

Use the answer choices. Step 3 shows each column-3 arrow can be reached in exactly $5$ ways, and how the bug finishes after that arrow doesn't depend on which of those $5$ ways it used. So the total is $5$ times some whole number, a multiple of $5$. Among the choices, only $2400$ ends in $0$ or $5$.