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# Six new employees,two of whom are married to each other,are to be assigned six desks that are lined up in a row.If the assignment of employees to desks is made randomly,what is the probability that the married couple will have non adjacent desks?

$\begin{array}{1 1}(A)\;\large\frac{4}{5}\\(B)\;\large\frac{9}{5}\\(C)\;\large\frac{2}{3}\\(D)\;\large\frac{2}{5}\end{array}$

Toolbox:
• Required probability =$\large\frac{n(E)}{n(S)}$
Given six employees
Two are married to each other.
Tital number of outcomes they can sit randomly in a row n(S)=6!
$\therefore$ Let us find the probability of adjacent desks.
Step 2:
There are two couples
$\therefore$ Number of outcomes of random arrangement of employees =5!
Consider 1 couple as 1 unit and arrangement of couple =2!
(ie) if a couple is AB then the seating arrangement can be AB or BA.
$\therefore$ Number of favorable outcomes n(E)=$5!\times 2!$
Step 3:
Required probability =$\large\frac{n(E)}{n(S)}$
$\Rightarrow \large\frac{5!\times 2!}{6!}$
$\Rightarrow \large\frac{2\times 1}{6}$
$\Rightarrow \large\frac{ 1}{3}$
Step 4:
$\therefore$ Probability of non-adjacent desks =1-probability of adjacent desks
$\Rightarrow 1-\large\frac{1}{3}$
$\Rightarrow \large\frac{2}{3}$
Hence (C) is the correct answer.