2011 AIME I Problems/Problem 12
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.
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:
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
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