Question

A certain number of boys and girls can be seated in a row such that no two girls are together in $1440$ ways. If one more boy joins them, the number of ways in which they can be seated in a row such that no two girls are together increases

• A. $4-fold$
• B. $6-fold$
• C. $8-fold$
• D. $10-fold$

Answer 1

The correct option is D $10-fold$

Let the no.of boys ${=}^{\prime }{x}^{\prime }$np. of girls ${=}^{\prime }{y}^{\prime }$If no $2$ girls are together, then $x+1\ge y$

Since $1$ more boy joins so:

No: ways to arrange him $x+1C_y$

$x!y!\frac{(x+1)!}{y!(x-y+1)!}=\frac{x!(x+1)!}{(x-y+1)!}=1440$

$1440$ is divisible $5$, $x+1\ge 5$

So $(x,y)$ can be $(3,4)$

On adding that $1$ rupees boy'

$\Rightarrow \frac{(x+1)!(x+2)!}{(x-y+2)!}=\frac{x!(x+1)!(x+1)(x+2)}{(x-y+1)!x-y+2)}$

$=\mathrm{1440.5.63}$

$\Rightarrow 1440\left(10\right)$

$=10$ fold

Option $D$ is correct