There are 5 students in class 10,6 students in class 11 and 8 students in class 12. If the number of ways, in which 10 students can be selected from them so as to include at least 2 students from each class and at most 5 students from the total 11 students of class 10 and 11 is 100k, then k is equal to

$$\begin{gathered} \mathrm{Number} \mathrm{of} {10}^{\mathrm{th}} \mathrm{class} \mathrm{students} = 5. \\ \mathrm{number} \mathrm{of} {11}^{\mathrm{th} }\mathrm{class} \mathrm{students} =6. \\ \mathrm{number} \mathrm{of} {12}^{\mathrm{th} }\mathrm{class} \mathrm{students} =8. \\ \mathrm{number} \mathrm{of} \mathrm{students} \mathrm{to} \mathrm{be} \mathrm{selected} =10 \\ \mathrm{and} \mathrm{atleast} 2 \mathrm{students} \mathrm{from} \mathrm{each} \mathrm{section} \\ \mathrm{atmost} 5 \mathrm{students} \mathrm{from} {10}^{\mathrm{th} }\mathrm{and} {11}^{\mathrm{th}} \end{gathered}$$

$$\begin{gathered} (5{\mathrm{C}}_2\times 6{\mathrm{C}}_{2}8{\mathrm{C}}_6)+(5{\mathrm{C}}_2\times 6{\mathrm{C}}_3\times 8{\mathrm{C}}_5)+(5{\mathrm{C}}_3\times 6{\mathrm{C}}_2\times 8{\mathrm{C}}_5) \\ =10\times 15\times 28+10\times 20\times 56+10\times 15\times 56 \\ =4200+11200+8400=23,800 \\ \therefore 23,800=100\mathrm{K} \Rightarrow \mathrm{k}=238 \end{gathered}$$