Standing in a Line in 2015

A group of 5 5 men and some women stand in a line. The probability that all 5 5 men stand next to each other, given that no man stands directly in between two women, is not greater than 2.015 % . 2.015\%. What is the least number of women that satisfies this?

Image source: Kluger Zoltan Wikimedia Commons


The answer is 49.

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1 solution

Steven Yuan
Jan 20, 2015

Let w w be the number of women. Also, let M M represent a man, W W represent a woman, and \underline{\; \;} be a line of some number of women (possibly 0). We have three cases:

M M M W M M M M W M M M M M M M M \underline{\: \:} MMMW \underline{\: \:} MM \underline{\: \:} \\ \underline{\: \:} MMW \underline{\: \:} MMM \underline{\: \:} \\ \underline{\: \:} MMMMM \underline{\: \:}

All three cases cover the possible scenarios. In our first case, if we treat men and women as indistinguishable, we have, from starts and bars, ( w + 1 2 ) \dbinom{w + 1}{2} possibilities for placing the women. The same can be said about our second case. Our third case, which is our desired case, has ( w + 1 1 ) \dbinom{w + 1}{1} ways to place the women. Thus, our probability that the five men stand next to each other is

( w + 1 1 ) 2 ( w + 1 2 ) + ( w + 1 1 ) = w + 1 w ( w + 1 ) + ( w + 1 ) = w + 1 ( w + 1 ) 2 = 1 w + 1 . \begin{aligned} \frac{\binom{w + 1}{1}}{2\binom{w + 1}{2} + \binom{w + 1}{1}} &= \frac{w + 1}{w(w + 1) + (w + 1)} \\ &= \frac{w + 1}{(w + 1)^2} \\ &= \frac{1}{w + 1}. \end{aligned}

We are given that this probability is no more than 0.02015 = 403 20000 . 0.02015 = \dfrac{403}{20000}. Solving this inequality gives w 48.627 , w \geq 48.627\dots, so w w is at least 49 . \boxed{49}.

You are forgetting the cases where a man is standing at an end of the row. (This would also satisfy your condition that no man is standing directly between two women.)

Jon Haussmann - 5 years, 5 months ago

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