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Hint: - Number of ways to sit along a circle by $n$ persons is$\left( {n - 1} \right)!$

Number of ways to sit along a line by n boys$ = n! $.

And the number of ways to sit along a line by n girls$ = n! $.

$\therefore $Starting from boy the number of ways to sit along a line by $n$ boys and $n$ girls alternately is $n! \times n! $.

Now, starting from girls, the number of ways to sit along a line by $n$ boys and $n$ girls alternately is $n! \times n! $.

Therefore total number of ways to sit along a line by $n$ boys and $n$ girls alternately

$

\left( x \right) = n! \times n! + n! \times n! \\

\Rightarrow x = 2 \times n! \times n! \\

$

Now, in circle starting does not matter because in the circle there are no starting and end points.

Therefore total no of ways to sit along a circle by $n$ boys and $n$ girls alternately

$ \Rightarrow y = \left( {n - 1} \right)! \times n! $

Now according to question it is given that $x = 12y$

$

\Rightarrow 2 \times n! \times n! = 12 \times \left( {n - 1} \right)! \times n! \\

\Rightarrow n! = 6 \times \left( {n - 1} \right)! \\

$

As we know that $n! = n\left( {n - 1} \right)!$

$

\Rightarrow n\left( {n - 1} \right)! = 6 \times \left( {n - 1} \right)! \\

\Rightarrow n = 6 \\

$

Hence, $n = 6$is the required answer.

$\therefore $Option (a) is correct.

Note: -In such types of questions first find out the total number of ways to sit along a line by $n$ boys and $n$ girls alternately and total number of ways to sit along a circle by $n$ boys and $n$ girls alternately, then equate them according to given condition then, we will get the required answer.

Number of ways to sit along a line by n boys$ = n! $.

And the number of ways to sit along a line by n girls$ = n! $.

$\therefore $Starting from boy the number of ways to sit along a line by $n$ boys and $n$ girls alternately is $n! \times n! $.

Now, starting from girls, the number of ways to sit along a line by $n$ boys and $n$ girls alternately is $n! \times n! $.

Therefore total number of ways to sit along a line by $n$ boys and $n$ girls alternately

$

\left( x \right) = n! \times n! + n! \times n! \\

\Rightarrow x = 2 \times n! \times n! \\

$

Now, in circle starting does not matter because in the circle there are no starting and end points.

Therefore total no of ways to sit along a circle by $n$ boys and $n$ girls alternately

$ \Rightarrow y = \left( {n - 1} \right)! \times n! $

Now according to question it is given that $x = 12y$

$

\Rightarrow 2 \times n! \times n! = 12 \times \left( {n - 1} \right)! \times n! \\

\Rightarrow n! = 6 \times \left( {n - 1} \right)! \\

$

As we know that $n! = n\left( {n - 1} \right)!$

$

\Rightarrow n\left( {n - 1} \right)! = 6 \times \left( {n - 1} \right)! \\

\Rightarrow n = 6 \\

$

Hence, $n = 6$is the required answer.

$\therefore $Option (a) is correct.

Note: -In such types of questions first find out the total number of ways to sit along a line by $n$ boys and $n$ girls alternately and total number of ways to sit along a circle by $n$ boys and $n$ girls alternately, then equate them according to given condition then, we will get the required answer.

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