Thank you for reporting, we will resolve it shortly
Q.
$5$ boys and $5$ girls are sitting in a row randomly. The probability that boys and girls sit alternatively is
Probability
Solution:
$\times B \times B \times B \times B \times B \times$
Since $5$ boys and $5$ girls sit alternately so there are $6$ places on which girls can be seated. If boys are arranged first (as shown in $'\times' $ sign)
So required probability $= \frac{^{5}P_{5} \times^{6}P_{5}}{^{10}P_{10}}$
$= \lfloor5 \times\frac{\lfloor6}{\lfloor10} $
$=\frac{1}{42}$