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Difference between revisions of "2001 AMC 12 Problems/Problem 16"

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Each dressing sequence can be uniquely described by a sequence containing two <math>1</math>s, two <math>2</math>s, ..., and two <math>8</math>s -- the first occurrence of number <math>x</math> means that the spider puts the sock onto leg <math>x</math>, the second occurrence of <math>x</math> means he puts the shoe onto leg <math>x</math>.  
 
Each dressing sequence can be uniquely described by a sequence containing two <math>1</math>s, two <math>2</math>s, ..., and two <math>8</math>s -- the first occurrence of number <math>x</math> means that the spider puts the sock onto leg <math>x</math>, the second occurrence of <math>x</math> means he puts the shoe onto leg <math>x</math>.  
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The number of such sequences is <math>{16\choose 2,2,\dots,2} = \frac{16!}{(2!)^8} = \boxed{\frac {16!}{2^8}}</math>.
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IF the numbers were all unique, the answer would be <math> 16! </math>. However, since 8 terms appear twice, the answer is <math> \frac{16!}{(2!)^8} = \boxed{\frac {16!}{2^8}}</math>.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2001|num-b=15|num-a=17}}
 
{{AMC12 box|year=2001|num-b=15|num-a=17}}

Revision as of 22:42, 20 January 2012

Problem

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?

$\text{(A) }8! \qquad \text{(B) }2^8 \cdot 8! \qquad \text{(C) }(8!)^2 \qquad \text{(D) }\frac {16!}{2^8} \qquad \text{(E) }16!$

Solution

Solution 1

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 $\boxed{\frac {16!}{2^8}}$.

Solution 2

Each dressing sequence can be uniquely described by a sequence containing two $1$s, two $2$s, ..., and two $8$s -- the first occurrence of number $x$ means that the spider puts the sock onto leg $x$, the second occurrence of $x$ means he puts the shoe onto leg $x$. IF the numbers were all unique, the answer would be $16!$. However, since 8 terms appear twice, the answer is $\frac{16!}{(2!)^8} = \boxed{\frac {16!}{2^8}}$.

See Also

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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