AMC8 2013

AMC8 2013 · Q21

AMC8 2013 · Q21. It mainly tests Basic counting (rules of product/sum), Combinations.

Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner of City Park, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take?

Samantha 住在城市公园西南角以西 2 个街区和以南 1 个街区的地方。她的学校位于城市公园东北角以东 2 个街区和以北 2 个街区的地方。在上学日，她骑自行车沿着街道到达城市公园的西南角，然后穿过公园的斜对角路径到达东北角，然后再骑自行车沿着街道去学校。如果她的路线尽可能短，她有多少种不同的路线可以选择？

(A) 3 3

(B) 6 6

(D) 12 12

(E) 18 18

Answer

Correct choice: (E)

正确答案：（E）

Solution

Using combinations, we get that the number of ways to get from Samantha's house to City Park is $\binom31 = \dfrac{3!}{1!2!} = 3$, and the number of ways to get from City Park to school is $\binom42= \dfrac{4!}{2!2!} = \dfrac{4\cdot 3}{2} = 6$. Since there's one way to go through City Park (just walking straight through), the number of different ways to go from Samantha's house to City Park to school $3\cdot 6 = \boxed{\textbf{(E)}\ 18}$.

使用组合数学，从 Samantha 家到城市公园的方式数是 $\binom{3}{1} = \dfrac{3!}{1!2!} = 3$，从城市公园到学校的方式数是 $\binom{4}{2} = \dfrac{4!}{2!2!} = \dfrac{4\cdot 3}{2} = 6$。穿过公园只有一种方式（直接走直线），因此从家到公园再到学校的不同方式总数是 $3\cdot 6 = \boxed{\textbf{(E)}\ 18}$。

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