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

Question

Six boys and six girls sit along a line alternatively in x ways and along a circle (again alternately in y ways), then (A) x = y (B) y = 12x (C) x = 10y (D) x = 12y

Tardigrade Admin · Accepted Answer

Clearly x =6!×6!+6!×6!=2(6 !)2 y=5!×6! ∴ (x/y)=(2(6!)2/5! 6!)=(2(6 !)/5 !)=(2×6(5!)/5!)=(12/1) ∴ x=12 y

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

$x = y$

$y = 12 x$

$x = 10 y$

$x = 12 y$

Clearly $x = 6! \times 6! + 6! \times 6! = 2 (6!)^{2}$
$y = 5! \times 6!$
$∴ \frac{x}{y} = \frac{2 ( 6 ! ) ^{2}}{5 ! 6 !} = \frac{2 ( 6 ! )}{5 !} = \frac{2 \times 6 ( 5 ! )}{5 !} = \frac{12}{1}$
$∴ x = 12 y$

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

x=y

y=12x

x=10y

x=12y