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

Question

Six boys and six girls sit along a line alternately 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

6 ! ⋅ 2 ⋅ 6 !=x, 5 ! 6 !=y ⇒ (x/y)=(6 ! ⋅ 2 ⋅ 6 !/5 ! ⋅ 6 !)=12 . Hence x=12 y ⇒ D

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

x = y

y = 12x

x = 10y

x = 12y

$6! \cdot 2 \cdot 6! = x, 5! 6! = y \Rightarrow \frac{x}{y} = \frac{6 ! \cdot 2 \cdot 6 !}{5 ! \cdot 6 !} = 12.$
Hence $x = 12 y \Rightarrow D$

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

x = y

y = 12x

x = 10y

x = 12y

6!⋅2⋅6!=x,5!6!=y⇒yx​=5!⋅6!6!⋅2⋅6!​=12. Hence x=12y⇒D

$6! \cdot 2 \cdot 6! = x, 5! 6! = y \Rightarrow \frac{x}{y} = \frac{6 ! \cdot 2 \cdot 6 !}{5 ! \cdot 6 !} = 12.$
Hence $x = 12 y \Rightarrow D$