2011 AIME I Problems/Problem 12

Revision as of 11:03, 2 March 2016 by Hesa57 (talk | contribs) (Solution)


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.


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:


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

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

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


Thus, we have

$\dfrac{  2\dbinom{n+1}{2} + \dbinom{n+1}{1}}{\dbinom{n+3}{3}+\dbinom{n+2}{2}}$

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 (ProblemsAnswer KeyResources)
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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