Question

一条 $16$ 步路径要从 $(-4,-4)$ 走到 $(4,4)$ ，每一步都使 $x$ 坐标或 $y$ 坐标增加 $1$ 。有多少条这样的路径在每一步都位于正方形 $-2\le x\le2$ 、 $-2\le y\le2$ 的外部或边界上？

A $16$ -step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$ -coordinate or the $y$ -coordinate by $1.$ How many such paths stay outside or on the boundary of the square $-2\le x\le2,$ $-2\le y\le2$ at each step?

$92$

$144$

$1568$

$1698$

$12{,}800$

答案：D

知识点：格路组合对称性

难度评级：1880

解答：

每一步都使 $x+y$ 增加 $1$ ，而坐标和从 $-8$ 变到 $8$ ，所以每条路径恰好经过一个满足 $x+y=0$ 的格点。

为了不进入开正方形内部，该点 $(t,-t)$ 必须满足 $|t|\ge2$ ，所以它是 $(\pm2,\mp2),(\pm3,\mp3),(\pm4,\mp4)$ 之一。

由对称性，只考虑三点 $(-4,4),(-3,3),(-2,2)$ ，然后将结果加倍。从 $(-4,-4)$ 到 $(-(4-j),4-j)$ 的路径数为 $\binom{8}{j}$ ，继续到 $(4,4)$ 的路径数也是 $\binom{8}{j}$ 。

因此 $\begin{gathered} 2\left(\binom80^2+\binom81^2+\binom82^2\right) \\ =2(1+64+784) \\ =1698. \end{gathered}$

所以正确答案是 D。

Every step increases $x+y$ by $1,$ which runs from $-8$ to $8,$ so each path passes through exactly one lattice point with $x+y=0.$

To stay out of the open square, that point $(t,-t)$ must have $|t|\ge2,$ so it is one of $(\pm2,\mp2),(\pm3,\mp3),(\pm4,\mp4).$

By symmetry consider the three points $(-4,4),(-3,3),(-2,2)$ and double. The number of paths from $(-4,-4)$ to $(-(4-j),4-j)$ is $\binom{8}{j},$ and the number continuing on to $(4,4)$ is also $\binom{8}{j}.$

Therefore the total is $\begin{gathered} 2\left(\binom80^2+\binom81^2+\binom82^2\right) \\ =2(1+64+784) \\ =1698. \end{gathered}$

Thus, D is the correct answer.

2010 AMC 12A 第 18 题

答案：D

难度评级：1880

解答：

其他年份的第 18 题