Question

$5$ boys and $5$ girls are sitting in a row randomly. The probability that boys and girls sit alternatively is

• A. $\frac{5}{126}$
• B. $\frac{1}{42}$
• C. $\frac{4}{126}$
• D. $\frac{6}{126}$

Answer 1

Answer: (B)

$\times B\times B\times B\times B\times B\times$
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 ${}^{\prime }{\times }^{\prime }$ sign)
So required probability $=\frac{{}^5{P}_5{\times }^6{P}_5}{{}^{10}{P}_{10}}$
$=\lfloor 5\times \frac{\lfloor 6}{\lfloor 10}$
$=\frac{1}{42}$