There are 5 students in class 10,6… - JEE Main 2021 Mathematics PYQ

JEE Main 2021 — Mathematics Algebra

2021integereasy

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

Official previous-year question

Held on 25 Jul 2021 · Verified 6 Jul 2026.

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Solution

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

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

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

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

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

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

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

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

$=23800$

According to question $100k=23800$

$\Rightarrow k=238$

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