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  4. Six boys and six girls sit along a line alternatively in x ways and along a circle (again alternately in y ways), then

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Q. Six boys and six girls sit along a line alternatively in $x$ ways and along a circle (again alternately in $y$ ways), then

Permutations and Combinations

Solution:

Clearly $x =6!\times6!+6!\times6!=2\left(6 !\right)^{2}$
$y=5!\times6!$
$\therefore \frac{x}{y}=\frac{2\left(6!\right)^{2}}{5!\,6!}=\frac{2\left(6 !\right)}{5 !}=\frac{2\times6\left(5!\right)}{5!}=\frac{12}{1}$
$\therefore x=12 \,y$