Ten students are seated at random in a row. The probability that two particular students are not seated side by side is

The correct option is A

$\frac{4}{5}$

Explanation for the correct option:

Find the probability that two particular students are not seated side by side:

Number of ways of seating ten students in a row $=10!$

Let $A$ and $B$ be two particular students.

Two students $A$ and $B$ can be seated side by side in two ways $A$$B$ or $B$$A$

Then the number of ways of two students out of ten seated side by side is $9!+9!=2\times 9!$

Hence, the probability of two students side by side$=\frac{\text{number of ways two student seated side by side}}{\text{total number of ways}}$

$$\begin{gathered} =\frac{2\times 9!}{10!} \\ =\frac{2}{10} \end{gathered}$$

Therefore, the probability of two students not seated side by side$=1-\text{probability of two students side by side}$

$$\begin{gathered} =1-\frac{2}{10} \\ =\frac{8}{10} \\ =\frac{4}{5} \end{gathered}$$

Hence, option (A) is the correct answer.