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

m (→‎Solution)
m (Multiple grammatical fixes based on subject-verb number agreement and several other corrections.)
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== Solution ==
 
== Solution ==
−
Denote (n) be n consecutive men and _ between (n) and (m) be some number of women between the mens(it can be zero).
+
Denote (n) be n consecutive men and _ between (n) and (m) be some number of women between the men (possibly zero).
  
 
There are five cases to consider:
 
There are five cases to consider:
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  _(6)_
 
  _(6)_
  
−
The first two cases gives 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 cases, we get <math>\dbinom{n+2}{2}</math> cases.
+
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.
  
−
The third to fifth cases counts the cases we desires.  
+
The third to fifth cases count the cases we desire.  
 
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.
 
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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so the probability is <math>\dfrac{  2\dbinom{n+1}{2} + \dbinom{n+1}{1}}{\dbinom{n+3}{3}+\dbinom{n+2}{2}}</math>
 
so the probability is <math>\dfrac{  2\dbinom{n+1}{2} + \dbinom{n+1}{1}}{\dbinom{n+3}{3}+\dbinom{n+2}{2}}</math>
  
−
the numerator simplifies into <math>(n+1)^2</math>.  
+
the numerator simplifies to <math>(n+1)^2</math>.  
  
−
The denominator simplifies into <math>\dfrac{(n+6)(n+2)(n+1)}{6}</math>
+
The denominator simplifies to <math>\dfrac{(n+6)(n+2)(n+1)}{6}</math>
  
−
so the whole faction simplifies into <math>\dfrac{6(n+1)}{(n+6)(n+2)}</math>
+
so the whole faction simplifies to <math>\dfrac{6(n+1)}{(n+6)(n+2)}</math>
  
−
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 equals to each other when <math>n = 594</math>.
+
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>.
  
−
If we let <math>n= 595</math> or <math>593</math>, we will notices that the answer is <math>\boxed{594}</math>
+
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}}

Revision as of 19:52, 3 April 2011

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 everyman 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

Denote (n) be n consecutive men and _ between (n) and (m) be some number of women between the men (possibly zero).

There are five cases to consider: 

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

_(3)_(3)_

_(2)_(4)_

_(4)_(2)_

_(6)_

The first two cases give us all the possible ways to arrange the people. Let there be $n$ women. For the first case, if we think of (n) as dividers, we get $\dbinom{n+3}{3}$ ways. For the second case, we get $\dbinom{n+2}{2}$ cases.

The third to fifth cases count the cases we desire. The third and fourth cases give us $2\dbinom{n+1}{2}$ if we put 1 woman between (2) and (4) before we count.

the last case gives us $\dbinom{n+1}{1}$

so the probability is $\dfrac{ 2\dbinom{n+1}{2} + \dbinom{n+1}{1}}{\dbinom{n+3}{3}+\dbinom{n+2}{2}}$

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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