Mathematics - Permutation & Combination Question with Solution | TestHub

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

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 ${}^{5}C_2\times {}^{6}C_3\times {}^{8}C_5$

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 $={}^{5}C_2\times {}^{6}C_2\times {}^{8}C_6$

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  ${}^{5}C_3\times {}^{6}C_2\times {}^{8}C_5$

$\Rightarrow$ Total number of ways $=\left({}^{5}C_2\times {}^{6}C_3\times {}^{8}C_5\right)+\left({}^{5}C_2\times {}^{6}C_2\times {}^{8}C_6\right)+\left({}^{5}C_3\times {}^{6}C_2\times {}^{8}C_5\right)$ $=\left({}^{5}C_2\times {}^{6}C_3\times {}^{8}C_3\right)+\left({}^{5}C_2\times {}^{6}C_2\times {}^{8}C_2\right)+\left({}^{5}C_2\times {}^{6}C_2\times {}^{8}C_3\right)$

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

$=23800$

According to question $100k=23800$

$\Rightarrow k=238$