Question

$6$ boys and $6$ girls are sitting in a row randomly. The probability that boys and girls sit alternately is:

• A. $\frac{1}{36}$
• B. $\frac{1}{462}$
• C. $\frac{5}{126}$
• D. $\frac{1}{231}$

Answer 1

The correct option is B

$\frac{1}{462}$

Explanation for the correct option:
Finding the required probability:

Let $S$ be the no. of ways in which $6$ boys and $6$ girls can sit in a row.

$\Rightarrow n\left(S\right)=12!$

Let $A$ be the event that the boys and girls sit alternately, this can be done in either of the $2$ ways:

$\mathrm{BG}\mathrm{BG}\mathrm{BG}\mathrm{BG}\mathrm{BG}\mathrm{BG}$

or

$\mathrm{GB}\mathrm{GB}\mathrm{GB}\mathrm{GB}\mathrm{GB}\mathrm{GB}$

In each way, the girls and boys can be seated in $6!$ ways each.

$\Rightarrow n\left(A\right)=2\times 6!\times 6!$

Therefore, the required probability is

$\begin{array}{rcl}P\left(A\right)& =& \frac{n\left(A\right)}{n\left(S\right)}\\ & =& \frac{2\times 6!\times 6!}{12!}\\ & =& \frac{2\times 6\times 5\times 4\times 3\times 2\times 1\times \cancel{6!}}{12\times 11\times 10\times 9\times 8\times 7\times \cancel{6!}}\\ & =& \frac{1}{462}\end{array}$

Hence, the correct answer is option B.