Indian Institute of Foreign Trade 2016 Solved Paper

(d) 5 students out of 4 boys and y girls can be chosen in 2 ways.
There are two conditions:
(i) 3 Boys & 2 Girls ${=}^4{\mathit{C}}_3{\times }^{\mathit{y}}{\mathit{C}}_2$
(ii) 4 Boys & 1 Girl${=}^4{\mathit{C}}_4{\times }^{\mathit{y}}{\mathit{C}}_1$
Since only boys are given a ball, thus
Total balls given to 3 boys each in 1${}^{\mathrm{st}}$ case + Total balls given to 4 boys each in 2${}^{\mathrm{nd}}$ case = 368
$\Rightarrow 3\times {}^4{\mathit{C}}_3\times {}^{\mathit{y}}{\mathit{C}}_2+4\times {}^4{\mathit{C}}_4\times {}^{\mathit{y}}{\mathit{C}}_1=368$
$\Rightarrow 3\times 4\times \frac{\mathit{y}\left(\mathit{y}-1\right)}{2}+4\times 1\times \mathit{y}=368$
$\Rightarrow 6\mathit{y}\left(\mathit{y}-1\right)+4\mathit{y}=368$
$\Rightarrow 6{\mathit{y}}^2-6\mathit{y}+4\mathit{y}=368$
$\Rightarrow 3{\mathit{y}}^2-\mathit{y}=184$
$\Rightarrow 3{\mathit{y}}^2-\mathit{y}-184=0$
on solving this. then, $\mathit{y}$ = 8.