The class contains $5$ rows of seats & each row having $8$ seats
∴ The classroom have a total of $(5 \times 8) = 40$ seats.
& the class has a total of $30$ students.
Now, the $6^{th}$ seat of row no. $5$ has to be empty.
& We need to find the probability of the same.
We know, $\text{Probability} = \dfrac{\text{No. of favorable ways}}{\text{Total no. of ways}}$
Now, the favorable ways will be when
$30$ students have the option to seat in $39$ seats. (Because $6^{th}$ seat of row no. $5$ needs to be empty)
∴ Out of $39$ seats $30$ students can seat in $^{39}C_{30}$ ways. [∵ As $30$ students have to choose $30$ seats out of $39$ seats.]
And Total no. of ways will be when
$30$ students have the option to seat in $40$ seats.
∴ No. of ways will be = $^{40}C_{30}$ways
$\color{violet}{\text{Now, the probability of the}}$ $\color{violet}{6^{th}}$ $\color{violet}{\text{ no. seat of row no.}}$ $\color{violet}{5}$ $\color{violet}{\text{ will left vacant is}}$ = $\dfrac{^{39}C_{30}}{^{40}C_{30}}$
$^{39}C_{30} = \dfrac{39 \times 38 \times .... \times 1}{(30 \times 29 \times .... \times 1) \times (9 \times 8 \times .... \times 1)}$
$\qquad \qquad = \dfrac{39 \times 38 \times .... \times 31}{9 \times 8 \times .... \times 1}$
$^{40}C_{30} = \dfrac{40 \times 39 \times .... \times 1}{(30 \times 29 \times .... \times 1) \times (10 \times 9 \times 8 \times .... \times 1)}$
$\qquad \qquad = \dfrac{40 \times 39 \times .... \times 31}{10 \times 9 \times 8 \times .... \times 1}$
∴$\dfrac{^{39}C_{30}}{^{40}C_{30}} =\dfrac{39 \times 38 \times .... \times 31}{9 \times 8 \times .... \times 1} \times \dfrac{10 \times 9 \times 8 \times .... \times 1}{40 \times 39 \times .... \times 31} $
$\qquad \qquad = \dfrac{10}{40}$
$\qquad \qquad = \color{maroon}{\dfrac{1}{4}}$
