Question

$5$ boys and $5$ girls sit in a row at random. The probability that the boys and girls an alternatively is

• A. $\frac{5}{14}$
• B. $\frac{3}{28}$
• C. $\frac{1}{126}$
• D. $\frac{1}{11}$

Answer 1

The correct option is C $\frac{1}{126}$

$5$ boys and $5$ girls sit in a row at random

$\therefore$ No of ways they can sit $=(5+5)!$

$=10!$

$\therefore$ No of ways that the boys and girls sit alternatively

$=51\times 5!\times 2$

$\therefore$ Probability that the boys and girls sit alternatively

$\frac{5!\times 5!\times 2}{10!}$

$=\frac{1}{126}$