Question

一只蜘蛛的八条腿各有一只袜子和一只鞋。若每条腿上必须先穿袜子再穿鞋，那么这只蜘蛛穿上所有袜子和鞋的顺序共有多少种？

A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe?

$8!$

$2^8\, 8!$

$(8!)^2$

$\dfrac{16!}{2^8}$

$16!$

答案：D

知识点：有限制的排列排列

难度评级：1600

小提示：

一共有 $16$ 件物品，若没有限制则有 $16!$ 种排列

There are $16$ items, so $16!$ orderings with no restriction

大提示：

对每条腿来说，在所有排列中袜子先于鞋子的排列恰占一半，八条腿的限制相互独立

On each leg the sock precedes the shoe in exactly half of all orderings, and the eight legs are independent

解答：

把 $16$ 件物品（ $8$ 只袜子和 $8$ 只鞋）按某个顺序排列：共有 $16!$ 种排列。

对每条腿，袜子在鞋子之前出现的排列恰好占所有排列的一半。对八条腿同时施加这个限制，需要除以 $2^8$ 得到 $\dfrac{16!}{2^8}\text{。}$

因此，正确答案是 D。

Think of the $16$ items ( $8$ socks and $8$ shoes) arranged in some order: there are $16!$ arrangements.

For each leg, the sock comes before the shoe in exactly half of all arrangements. Imposing this on all eight legs independently divides by $2^8,$ giving $\dfrac{16!}{2^8}.$

Thus, the correct answer is D.

2001 AMC 12 第 16 题

答案：D

难度评级：1600

小提示：

大提示：

解答：

其他年份的第 16 题