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. $\frac{1}{126}$

Answer 1

The correct option is D

$\frac{1}{126}$

Explanation for the correct option:

Step 1: Find the total sitting arrangments possible:

Since there are $10$ students in total.

So, number of ways in which they can sit $=10!$

Step 2: Find the number of favourable arrangments:

As there are $5$ boys and $5$ girls.

So, number of ways in which they can sit alternatively $2\times 5!\times 5!$

Step 3: Find the probability of alternative seating:

Divide Favourable arrangments by total arrangments to get the probability.

Hence, the required probability:

$$\begin{gathered} p=\frac{2\times 5!\times 5!}{10!} \\ =\frac{2\times 120}{10\times 9\times 8\times 7\times 6} \\ =\frac{1}{126} \end{gathered}$$

Hence, option (D) is the correct answer.