Help the Spider!

Probability Level 3

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?

This problem was taken from 2001 AMC 12 problems.

{ 2 }^{ 8 }\cdot 8!

{ (8!) }^{ 2 }

\frac { 16! }{ { 2 }^{ 8 } }

16!

2 solutions

Mateo Matijasevick
Apr 2, 2016

Let the spider try to put on all $16$ things in a random order. Each of the $16!$ permutations is equally probable. For any fixed leg, the probability that he will first put on the sock and only then the shoe is clearly $\frac { 1 }{ 2 }$ . Then the probability that he will correctly put things on all legs is $\frac { 1 }{ { 2 }^{ 8 } }$ . Therefore the number of correct permutations must be $\frac { 16! }{ { 2 }^{ 8 } }$

If it was given that he will always put on socks before shoes then what are the total permutations?

Siva Bathula - 5 years, 2 months ago

Prince Loomba
Nov 2, 2016

It can be considered as a permutation of

AABBCC.....HH

As each alphabet representing a particular leg!

So the answer is $\frac{16!}{2^{8}}$

Help the Spider!

This problem was taken from 2001 AMC 12 problems.

2 solutions

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