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

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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.
 
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 second, third, and fourth cases are like the first, only that 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> ways.
  
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The last case gives us <math>\dbinom{n+1}{1}</math> ways.
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The last case gives us <math>\dbinom{n+1}{1}=n+1</math> ways.
  
 
Therefore, the total number of possible ways where there are no isolated men is
 
Therefore, the total number of possible ways where there are no isolated men is
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<cmath>\dbinom{n+1}{3}+3\dbinom{n+1}{2}+(n+1).</cmath>
 
<cmath>\dbinom{n+1}{3}+3\dbinom{n+1}{2}+(n+1).</cmath>
  
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Thus, we have
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The total number of ways where there is a group of at least four men together is the sum of the third, fourth, and fifth case, or
  
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<math>\dfrac{  2\dbinom{n+1}{2} + \dbinom{n+1}{1}}{\dbinom{n+3}{3}+\dbinom{n+2}{2}}</math>
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<cmath>2\dbinom{n+1}{2}+(n+1).</cmath>
  
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the numerator simplifies to <math>(n+1)^2</math>.
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Thus, we want to find the minimum possible value of <math>n</math> where <math>n</math> is a positive integer such that
  
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The denominator simplifies to <math>\dfrac{(n+6)(n+2)(n+1)}{6}</math>
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<cmath>\dfrac{2\dbinom{n+1}{2}+(n+1)}{\dbinom{n+1}{3}+3\dbinom{n+1}{2}+(n+1)}\le\dfrac{1}{100}.</cmath>
  
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so the whole faction simplifies to <math>\dfrac{6(n+1)}{(n+6)(n+2)}</math>
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After simplification, we arrive at <cmath>\dfrac{6(n+1)}{n^2+8n+6}\le\dfrac{1}{100}.</cmath>
  
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Since <math>\dfrac{n+1}{n+2}</math> is slightly less than 1 when <math>n</math> is large, <math>\dfrac{6}{n+6}</math> will be close to <math>\dfrac{1}{100}</math>. They equal each other when <math>n = 594</math>.
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Simplifying again, we see that we seek the smallest positive integer value of <math>n</math> such that <math>n(n-592)\ge594</math>. Clearly <math>n>592</math>, or the left side will not even be positive; we quickly see that <math>n=593</math> is too small but <math>n=\boxed{594}</math> satisfies the inequality.
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If we let <math>n= 595</math> or <math>593</math>, we will notice that the answer is <math>\boxed{594}</math>
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2011|n=I|num-b=11|num-a=13}}
 
{{AIME box|year=2011|n=I|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:46, 17 February 2020

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 that we need to insert two dividers among the $n+1$ possible locations. Each gives us $\dbinom{n+1}{2}$ ways, for a total of $3\dbinom{n+1}{2}$ ways.

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

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

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

The total number of ways where there is a group of at least four men together is the sum of the third, fourth, and fifth case, or

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

Thus, we want to find the minimum possible value of $n$ where $n$ is a positive integer such that

\[\dfrac{2\dbinom{n+1}{2}+(n+1)}{\dbinom{n+1}{3}+3\dbinom{n+1}{2}+(n+1)}\le\dfrac{1}{100}.\]

After simplification, we arrive at \[\dfrac{6(n+1)}{n^2+8n+6}\le\dfrac{1}{100}.\]

Simplifying again, we see that we seek the smallest positive integer value of $n$ such that $n(n-592)\ge594$. Clearly $n>592$, or the left side will not even be positive; we quickly see that $n=593$ is too small but $n=\boxed{594}$ satisfies the inequality.

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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