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.