Question

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 $x:y=$

• A. 12 : 1
• B. 42 : 1
• C. 16 : 1
• D. 6 : 1

Answer 1

Answer: (B)

$6$ 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.
$\therefore x=6!\times 7_{P_6}=6!\times 7!$
$6$ Boys can be seated in a circle in $(6-1)!$ ways $=5!$
Now, in the $6$ gaps $6$ girls can be arranged in $6_{P_6}$ ways.
$\therefore y=5!\times 6_{P_6}=5!\times 6!$
Now, $x:y=6!\times 7!:5!\times 6!$
$\Rightarrow x:y=7!:5!$
$\Rightarrow x:y=7\times 6\times 5!:5!$
$\Rightarrow x:y=42:1$