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 !$
$\Rightarrow \, x: y=7 !: 5 !$
$\Rightarrow \,x: y=7 \times 6 \times 5 !: 5 !$
$\Rightarrow \, x: y=42: 1$