If $x$ is the number of ways in which six women and six men can be arranged to sit in a row such that no two…
If $x$ is the number of ways in which six women and six men can be arranged to sit in a row such that no two women are together and if $y$ is the number of ways they are seated around a table in the same manner, then $x: y=$
$12: 1$
$42: 1$
$16: 1$
$6: 1$
Solution
6 Boys can be seated in a row in $6_{P_6}$ ways $=6 !$. Now, in the 7 gaps 6 girls can be arranged in $7_{P_6}$ ways.
$
\therefore \quad x=6 ! \times 7_{P_6}=6 ! \times 7 !
$
6 Boys can be seated in a circle in
$
(6-1) \text { ! ways }=5 \text { ! }
$
Now, in the 6 gaps 6 girls can be arranged in $6_{P_6}$ ways.
$
\begin{array}{llrl}
\therefore & & y=5 ! \times 6_{P_6}=5 ! \times 6 ! \\
\text { Now, } & x: y=6 ! \times 7 !: 5 ! \times 6 ! \\
\Rightarrow & x: y=7 !: 5 ! \\
\Rightarrow & x: y=7 \times 6 \times 5 !: 5 ! \\
\Rightarrow & x: y=42: 1
\end{array}
$