Question

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?

$8!$

$2^8\, 8!$

$(8!)^2$

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

$16!$

Answer: D

Concepts:arrangements with restrictions permutations

Difficulty rating: 1600

Solution:

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 Problem 16

Answer: D

Difficulty rating: 1600

Solution:

Problem 16 in Other Years