Question

If n boys and n girls sit along a line alternately in x ways and along a circle alternately in y ways such that x = 12y then n is equal to:
$
  {\text{a}}{\text{. 6}} \\
  {\text{b}}{\text{. 8}} \\
  {\text{c}}{\text{. 9}} \\
  {\text{d}}{\text{. 12}} \\
$

Answer 1

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.