# 2011 AIME I Problems/Problem 12

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```== 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 everyman 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 a space between (n) and (m) be some number of women between the mens(it can be zero).

There are five cases to consider:

```(2) (2) (2)
```
```(3) (3)
```
```(2) (4)
```
```(4) (2)
```
```(6)
```

The first two cases gives us all the possible ways to arrange the people. Let there be $k$ women. For the first case, if we think of (n) as dividers, we get $\dbinom{n+3}{3}$ ways. For the second cases, we get $\dbinom{n+2}{2}$ cases.

The third to fifty cases counts the cases we desires. 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 into $(n+1)^2$.

The denominator simplifies into $\dfrac{(n+6)(n+2)(n+1)}{6}$

so the whole faction simplifies into $\dfrac{6(n+1)}{(n+6)(n+2)}$

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 equals to each other when $n = 594$.

If we let $n= 595$ or $593$, we will notices that the answer is \$\boxed{594}