men and some women stand in a line. The probability that all men stand next to each other, given that no man stands directly in between two women, is not greater than What is the least number of women that satisfies this?
A group of
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Let w be the number of women. Also, let M represent a man, W represent a woman, and 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
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, ( 2 w + 1 ) possibilities for placing the women. The same can be said about our second case. Our third case, which is our desired case, has ( 1 w + 1 ) ways to place the women. Thus, our probability that the five men stand next to each other is
2 ( 2 w + 1 ) + ( 1 w + 1 ) ( 1 w + 1 ) = w ( w + 1 ) + ( w + 1 ) w + 1 = ( w + 1 ) 2 w + 1 = w + 1 1 .
We are given that this probability is no more than 0 . 0 2 0 1 5 = 2 0 0 0 0 4 0 3 . Solving this inequality gives w ≥ 4 8 . 6 2 7 … , so w is at least 4 9 .