JEE Main 2021 — Mathematics Algebra

$$\begin{matrix}\mathrm{Class} & {10}^{\mathrm{th}} & {11}^{\mathrm{th}} & {12}^{\mathrm{th}} \\ \mathrm{Total}\mathrm{student} & 5 & 6 & 8\end{matrix}$$ClassTotalstudent​10th5​11th6​12th8​

From these students, we can choose two students from class $10$10, three students from class 11&5 students from class $12.$12.

So, total number of ways is $C25\times C36\times C58$C25×C36×C58

Again, From these students, we can choose two students from class $10,$10, two students from class 11&6 students from class $12.$12.

Number of selection $=C25\times C26\times C68$=C25×C26×C68

Again, from these students, we can choose three students from class $10$10, two students from class 11&5 students from class $12.$12. So, total number of ways is $C35\times C26\times C58$C35×C26×C58

$\Rightarrow$⇒ Total number of ways $=(C25\times C36\times C58)+(C25\times C26\times C68)+(C35\times C26\times C58)$=(C25×C36×C58)+(C25×C26×C68)+(C35×C26×C58) $=(C25\times C36\times C38)+(C25\times C26\times C28)+(C25\times C26\times C38)$=(C25×C36×C38)+(C25×C26×C28)+(C25×C26×C38)

$$=(\frac{5\times 4}{2\times 1}\times \frac{6\times 5\times 4}{3\times 2\times 1}\times \frac{8\times 7\times 6}{3\times 2\times 1})+(\frac{5\times 4}{2\times 1}\times \frac{6\times 5}{2\times 1}\times \frac{8\times 7}{2\times 1})+(\frac{5\times 4}{2\times 1}\times \frac{6\times 5}{2\times 1}\times \frac{8\times 7\times 6}{3\times 2\times 1})$$=(2×15×4​×3×2×16×5×4​×3×2×18×7×6​)+(2×15×4​×2×16×5​×2×18×7​)+(2×15×4​×2×16×5​×3×2×18×7×6​)

$$=23800$$=23800

According to question $100k=23800$100k=23800

$$\Rightarrow k=238$$⇒k=238