Difference between revisions of "2011 AIME I Problems/Problem 12"
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Revision as of 19:25, 4 July 2013
Problem
Six men and some number of women stand in a line in random order. Let 
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 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 
women. For the first case, if we think of (n) as dividers, we get 
ways. For the second case, we get 
cases.
The third to fifth cases count the cases we desire.
The third and fourth cases give us 
if we put 1 woman between (2) and (4) before we count.
the last case gives us 
so the probability is 
the numerator simplifies to 
.
The denominator simplifies to 
so the whole faction simplifies to 
Since 
is slightly less than 1 when is large, 
will be close to 
. They equal each other when 
.
If we let 
or 
, we will notice that the answer is 
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. 
