There are 5 students in class 10, 6 students in class 11 and 8...

$\begin{cases}Class & {10}^{th} & {11}^{th} & {12}^{th} \\ Totalstudent & 5 & 6 & 8\end{cases}$$</p><p>From these students, we can choose two students from class$10$, three students from class$11 \text{ and } 5$students from class$12 .$</p><p>So, total number of ways is$C2 5 \times C3 6 \times C5 8$</p><p>Again, From these students, we can choose two students from class$10,$two students from class$11 \text{ and } 6$students from class$12 .$</p><p>Number of selection$= C2 5 \times C2 6 \times C6 8$</p><p>Again, from these students, we can choose three students from class$10$, two students from class$11 \text{ and } 5$students from class$12 .$So, total number of ways is$C3 5 \times C2 6 \times C5 8$</p><p>$⇒$Total number of ways$= \left(C2 5 \times C3 6 \times C5 8\right) + \left(C2 5 \times C2 6 \times C6 8\right) + \left(C3 5 \times C2 6 \times C5 8\right)$= \left(C2 5 \times C3 6 \times C3 8\right) + \left(C2 5 \times C2 6 \times C2 8\right) + \left(C2 5 \times C2 6 \times C3 8\right)$

$= \left(\dfrac{5 \times 4}{2 \times 1} \times \dfrac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1}\right) + \left(\dfrac{5 \times 4}{2 \times 1} \times \dfrac{6 \times 5}{2 \times 1} \times \dfrac{8 \times 7}{2 \times 1}\right) + \left(\dfrac{5 \times 4}{2 \times 1} \times \dfrac{6 \times 5}{2 \times 1} \times \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1}\right)$

$= 23800$

According to question $100k = 23800$

$⇒ k = 238$