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Difference between revisions of "2011 AIME I Problems/Problem 12"

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== Solution ==
 
== Solution ==
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Denote (n) be n consecutive men and _ between (n) and (m) be some number of women between the men greater than zero.
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Let <math>n</math> be the number of women present, and let _ be some positive number of women between groups of men.  Since the problem states that every man stands next to another man, there cannot be isolated men. Thus, there are five cases to consider, where <math>(k)</math> refers to a consecutive group of <math>k</math> men:
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There are five cases to consider:
 
  
 
  _(2)_(2)_(2)_
 
  _(2)_(2)_(2)_
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  _(6)_
 
  _(6)_
  
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The first two cases give us all the possible ways to arrange the people. Let there be <math>n</math> women. For the first case, if we think of (n) as dividers, we get <math>\dbinom{n+3}{3}</math> ways. For the second case, we get <math>\dbinom{n+2}{2}</math> cases.
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For the first case, we can place the three groups of men in between women.  We can think of the groups of men as dividers splitting up the <math>n</math> women. Since there are <math>n+1</math> possible places to insert the dividers, and we need to choose any three of these locations, we have <math>\dbinom{n+1}{3}</math> ways.
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The second, third, and fourth cases are like the first, only we need to insert two dividers among the <math>n+1</math> possible locations.  Each gives us <math>\dbinom{n+1}{2}</math> ways, for a total of <math>3\dbinom{n+1}{2}</math>.
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The last case gives us <math>\dbinom{n+1}{1}</math> ways.
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Therefore, the total number of possible ways where there are no isolated men is
  
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The third to fifth cases count the cases we desire.
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<cmath>\dbinom{n+1}{3}+3\dbinom{n+1}{2}+(n+1).</cmath>
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The third and fourth cases give us <math>2\dbinom{n+1}{2}</math> if we put 1 woman between (2) and (4) before we count.
 
  
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the last case gives us <math>\dbinom{n+1}{1}</math>
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Thus, we have
  
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so the probability is <math>\dfrac{  2\dbinom{n+1}{2} + \dbinom{n+1}{1}}{\dbinom{n+3}{3}+\dbinom{n+2}{2}}</math>
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<math>\dfrac{  2\dbinom{n+1}{2} + \dbinom{n+1}{1}}{\dbinom{n+3}{3}+\dbinom{n+2}{2}}</math>
  
 
the numerator simplifies to <math>(n+1)^2</math>.  
 
the numerator simplifies to <math>(n+1)^2</math>.  

Revision as of 12:03, 2 March 2016

Problem

Six men and some number of women stand in a line in random order. Let $p$ be the probability that a group of at least four men stand together in the line, given that every man stands next to at least one other man. Find the least number of women in the line such that $p$ does not exceed 1 percent.

Solution

Let $n$ be the number of women present, and let _ be some positive number of women between groups of men. Since the problem states that every man stands next to another man, there cannot be isolated men. Thus, there are five cases to consider, where $(k)$ refers to a consecutive group of $k$ men: 

_(2)_(2)_(2)_

_(3)_(3)_

_(2)_(4)_

_(4)_(2)_

_(6)_

For the first case, we can place the three groups of men in between women. We can think of the groups of men as dividers splitting up the $n$ women. Since there are $n+1$ possible places to insert the dividers, and we need to choose any three of these locations, we have $\dbinom{n+1}{3}$ ways.

The second, third, and fourth cases are like the first, only we need to insert two dividers among the $n+1$ possible locations. Each gives us $\dbinom{n+1}{2}$ ways, for a total of $3\dbinom{n+1}{2}$.

The last case gives us $\dbinom{n+1}{1}$ ways.

Therefore, the total number of possible ways where there are no isolated men is

\[\dbinom{n+1}{3}+3\dbinom{n+1}{2}+(n+1).\]

Thus, we have

$\dfrac{ 2\dbinom{n+1}{2} + \dbinom{n+1}{1}}{\dbinom{n+3}{3}+\dbinom{n+2}{2}}$

the numerator simplifies to $(n+1)^2$.

The denominator simplifies to $\dfrac{(n+6)(n+2)(n+1)}{6}$

so the whole faction simplifies to $\dfrac{6(n+1)}{(n+6)(n+2)}$

Since $\dfrac{n+1}{n+2}$ is slightly less than 1 when $n$ is large, $\dfrac{6}{n+6}$ will be close to $\dfrac{1}{100}$. They equal each other when $n = 594$.

If we let $n= 595$ or $593$, we will notice that the answer is $\boxed{594}$

See also

2011 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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