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6 employees, 2 of them are married to each other, are to be assigned desks that are lined up in a row. if the desks are chosen at random, what is the probability that the married couple will have non adjacent desks?

First we need to find the probability of the married couple getting adjacent seats, and then subtract it from 1.

$letA,B,C,D,E,Fare6employeesandletAandBarecouple.\phantom{\rule{0ex}{0ex}}Nownumberofwaystoarrange6peoplekeepingABsidebyside=\phantom{\rule{0ex}{0ex}}Numberofwaystoarrange\left[AB\right],C,D,E,F+Numberofwaystoarrange\left[BA\right],C,D,E,F\phantom{\rule{0ex}{0ex}}Therefore\phantom{\rule{0ex}{0ex}}Nownumberofwaystoarrange6peoplekeepingABsidebyside\phantom{\rule{0ex}{0ex}}=5!+5!\phantom{\rule{0ex}{0ex}}=120+120\phantom{\rule{0ex}{0ex}}=240\phantom{\rule{0ex}{0ex}}Nowthenumberofwaystoarrange6peoplewillbemytotalsamplesize\phantom{\rule{0ex}{0ex}}=6!\phantom{\rule{0ex}{0ex}}=720\phantom{\rule{0ex}{0ex}}Henceprobabilitythatcouplewillhaveadjacentdesk\phantom{\rule{0ex}{0ex}}=\frac{240}{720}\phantom{\rule{0ex}{0ex}}P\left(r\right)=\frac{1}{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Theprobabilitythatcouplewillhavenonadjacentdesk.\phantom{\rule{0ex}{0ex}}1-P\left(r\right)\phantom{\rule{0ex}{0ex}}=1-\frac{1}{3}\phantom{\rule{0ex}{0ex}}Answer=\frac{2}{3}$

Regards,

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