AMC12 2012 B

AMC12 2012 B · Q22

AMC12 2012 B · Q22. It mainly tests Basic counting (rules of product/sum), Casework.

A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?

一只虫子沿着下图所示六角形格点中的线段从 A 移动到 B。标有箭头的线段只能沿箭头方向行走，且虫子不会重复走同一线段。有多少条不同的路径？

(A) 2112 2112

(B) 2304 2304

(D) 2384 2384

(E) 2400 2400

Answer

Correct choice: (E)

正确答案：（E）

Solution

There is $1$ way to get to any of the red arrows. From the first (top) red arrow, there are $2$ ways to get to each of the first and the second (top 2) blue arrows; from the second (bottom) red arrow, there are $3$ ways to get to each of the first and the second blue arrows. So there are in total $5$ ways to get to each of the blue arrows. From each of the first and second blue arrows, there are respectively $4$ ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively $8$ ways to get to each of the first and the second green arrows. Therefore there are in total $5 \cdot (4+4+8+8) = 120$ ways to get to each of the green arrows. Finally, from each of the first and second green arrows, there are respectively $2$ ways to get to the first orange arrow; from each of the third and the fourth green arrows, there are $3$ ways to get to the first orange arrow. Therefore there are $120 \cdot (2+2+3+3) = 1200$ ways to get to each of the orange arrows, hence $2400$ ways to get to the point $B$. $\boxed{\textbf{(E)}\ 2400}$

到达任意红色箭头处都只有 $1$ 种走法。从第一个（上方）红色箭头出发，到达第一个和第二个（上方两个）蓝色箭头各有 $2$ 种走法；从第二个（下方）红色箭头出发，到达第一个和第二个蓝色箭头各有 $3$ 种走法。因此到达每个蓝色箭头共有 $5$ 种走法。从第一个和第二个蓝色箭头出发，到达第一个和第二个绿色箭头分别各有 $4$ 种走法；从第三个和第四个蓝色箭头出发，到达第一个和第二个绿色箭头分别各有 $8$ 种走法。因此到达每个绿色箭头共有 $5\cdot(4+4+8+8)=120$ 种走法。最后，从第一个和第二个绿色箭头出发，到达第一个橙色箭头分别各有 $2$ 种走法；从第三个和第四个绿色箭头出发，到达第一个橙色箭头各有 $3$ 种走法。因此到达每个橙色箭头共有 $120\cdot(2+2+3+3)=1200$ 种走法，从而到达点 $B$ 共有 $2400$ 种走法。$\boxed{\textbf{(E)}\ 2400}$

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