# 2011 AIME I Problems/Problem 12

## 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

Denote (n) be n consecutive men and _ between (n) and (m) be some number of women between the men greater than 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 $\boxed{594}$