AMC12 2008 B

AMC12 2008 B · Q22

AMC12 2008 B · Q22. It mainly tests Combinations, Casework.

A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers chose spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 2 adjacent spaces. What is the probability that she is able to park?

一个停车场有一排 16 个车位。12 辆车到达，每辆车需要 1 个车位，司机从仍然空着的车位中随机选择。随后 Em 阿姨开着她的 SUV 到达，这辆车需要 2 个相邻车位。她能够停车的概率是多少？

(A) \(\dfrac{11}{20}\) \(\dfrac{11}{20}\)

(B) \(\dfrac{4}{7}\) \(\dfrac{4}{7}\)

(D) \(\dfrac{3}{5}\) \(\dfrac{3}{5}\)

(E) \(\dfrac{17}{28}\) \(\dfrac{17}{28}\)

Answer

Correct choice: (E)

正确答案：（E）

Solution

Auntie Em won't be able to park only when none of the four available spots touch. We can form a bijection between all such cases and the number of ways to pick four spots out of 13: since none of the spots touch, remove a spot from between each of the cars. From the other direction, given four spots out of 13, simply add a spot between each. So the probability she can park is \[1-\frac{{13 \choose 4}}{{16 \choose 4}}=1-\frac{13\cdot12\cdot11\cdot10}{16\cdot15\cdot14\cdot13}=1-\frac{11}{28}={\textbf{(E)}\frac{17}{28}}.\] (Bijection: When elements of two sets are perfectly paired with each other, i.e. each and every element from both sets has exactly one match in the other, and no elements are left out. In the context of this problem, this means the number of distinct ways to order the cars such that no two spaces are adjacent is exactly the number of ways to pick 4 spots out of 13.)

Em 阿姨无法停车当且仅当剩下的 4 个空位两两都不相邻。我们可以在这种情况与从 13 个位置中选 4 个位置之间建立一个双射：由于空位互不相邻，在每两个空位之间去掉一个位置；反过来，给定从 13 个位置中选出的 4 个位置，只需在每两个之间加回一个位置即可。因此她能停车的概率为 \[1-\frac{{13 \choose 4}}{{16 \choose 4}}=1-\frac{13\cdot12\cdot11\cdot10}{16\cdot15\cdot14\cdot13}=1-\frac{11}{28}={\textbf{(E)}\frac{17}{28}}.\] （双射：当两个集合的元素可以一一配对，即两个集合中每个元素在另一个集合中恰好有一个对应且无遗漏时，称为双射。在本题中，这意味着“选取 4 个互不相邻的空位”的不同方式数，恰好等于“从 13 个位置中选 4 个位置”的方式数。）

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