Question

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

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

Answer 1

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

Total number of ways $=10!$
Total number of ways in which $5$ boys and $5$ girls are sitting in a row $=2\times 5!\times 5!$
$\therefore$ Required probability
$=\frac{2\times 5!\times 5!}{10!}=\frac{2\times 5!\times 5\times 4\times 3\times 2}{10\times 9\times 8\times 7\times 6\times 5!}=\frac{1}{126}$