AMC12 2001 A

AMC12 2001 A · Q16

AMC12 2001 A · Q16. It mainly tests Basic counting (rules of product/sum), Permutations.

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?

一只蜘蛛为它的八条腿各有一只袜子和一只鞋。蜘蛛穿上袜子和鞋的不同顺序有多少种，假设每条腿上袜子必须在鞋子之前穿上？

(A) 8! 8!

(B) $2^8$! $2^8$!

(D) $\frac{16!}{2^8}$ $\frac{16!}{2^8}$

(E) 16! 16!

Answer

Correct choice: (D)

正确答案：（D）

Solution

Suppose the spider tries to put on all $2 \cdot 8 = 16$ items in a random order, so that each of the $16!$ possible permutations is equally probable. This means that for any fixed leg, the probability that it will first put on the sock, and only then the shoe, is clearly $\frac{1}{2}$. Hence the probability that it will put on the shoe and sock in the correct order for all its legs is $\left(\frac{1}{2}\right)^8 = \frac{1}{2^{8}}$. Therefore the number of possible permutations is $\boxed{\text{(D) }\frac {16!}{2^8}}$.

假设蜘蛛尝试以随机顺序穿上全部 $2\cdot 8 = 16$ 件物品，因此 $16!$ 种排列中的每一种都等可能。这意味着对任意固定的一条腿，先穿袜子再穿鞋的概率显然是 $\frac{1}{2}$。因此，它在所有腿上都按正确顺序穿鞋袜的概率是 $\left(\frac{1}{2}\right)^8 = \frac{1}{2^{8}}$。所以满足条件的排列数为 $\boxed{\text{(D) }\frac {16!}{2^8}}$。

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