Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (Multiple grammatical fixes based on subject-verb number agreement and several other corrections.)
(→‎Solution)
 
(25 intermediate revisions by 8 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
Six men and some number of women stand in a line in random order. Let <math>p</math> 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 <math>p</math> does not exceed 1 percent.
+
Six men and some number of women stand in a line in random order. Let <math>p</math> 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 <math>p</math> does not exceed 1 percent.
  
 
== Solution ==
 
== Solution ==
−
Denote (n) be n consecutive men and _ between (n) and (m) be some number of women between the men (possibly zero).
+
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:
−
 
 
−
There are five cases to consider:
 
  
 
  _(2)_(2)_(2)_
 
  _(2)_(2)_(2)_
Line 17: Line 15:
 
  _(6)_
 
  _(6)_
  
−
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.
+
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.
 +
 
 +
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.
 +
 
 +
The last case gives us <math>\dbinom{n+1}{1}=n+1</math> ways.
  
−
The third to fifth cases count the cases we desire.
+
Therefore, the total number of possible ways where there are no isolated men is
−
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 last case gives us <math>\dbinom{n+1}{1}</math>
+
<cmath>\dbinom{n+1}{3}+3\dbinom{n+1}{2}+(n+1).</cmath>
  
−
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 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
  
−
the numerator simplifies to <math>(n+1)^2</math>.
+
<cmath>2\dbinom{n+1}{2}+(n+1).</cmath>
  
−
The denominator simplifies to <math>\dfrac{(n+6)(n+2)(n+1)}{6}</math>
+
Thus, we want to find the minimum possible value of <math>n</math> where <math>n</math> is a positive integer such that
  
−
so the whole faction simplifies to <math>\dfrac{6(n+1)}{(n+6)(n+2)}</math>
+
<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>
  
−
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>.
+
After simplification, we arrive at <cmath>\dfrac{6(n+1)}{n^2+8n+6}\le\dfrac{1}{100}.</cmath>
  
−
If we let <math>n= 595</math> or <math>593</math>, we will notice that the answer is <math>\boxed{594}</math>
+
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.
  
 
== 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}}

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_12&oldid=117857"