In an examination, the maximum marks each of three papers is $50$ and the maximum mark for the fourth paper is $100.$ Find the number of ways in which the candidate can score $60\%$ marks in aggregate. 

Aggregate of marks $=50\times 3+100=250$

$\therefore 60\%$ of the aggregate $=\frac{60}{100}\times 250=150$

Let the marks scored by the candidate in four papers be $x_1, x_2 , x_3$ and $x_4$. Here, the number of required ways will be equal to the number of Sols of .$x_1 + x_2 + x_3 + x_4 = 150$ i.e., $0\le x_1, x_2, x_3\le 50$and $0\le x_4\le 100.$

 Since, the upper limit is $100 <$ required sum $\left(150\right).$ 

The number of solutions of the equation is equal to coeffcient of ${\alpha }^{150}$expansion of 

  ${\left({\alpha }^0+{\alpha }^1+{\alpha }^2+\dots +{\alpha }^{50}\right)}^3\left({\alpha }^0+{\alpha }^1+{\alpha }^2+\dots +{\alpha }^{100}\right)$

$=$Coefficient of ${\alpha }^{150}$ in the expansion of ${\left(1-{\alpha }^{51}\right)}^3\left(1-{\alpha }^{10}\right)(1-\alpha {)}^{-1}$

$=$Coefficient of ${\alpha }^{150}$ in the expansion of $\left(1-3{\alpha }^{51}+3{\alpha }^{102}\right)\left(1-{\alpha }^{101}\right)\left(1{+}^4{C}_1\alpha {+}^5{C}_2{\alpha }^2+\dots +\mathrm{\infty }\right)$

$=$Coefficient of ${\alpha }^{150}$ in the expansion of $\left(1-3{\alpha }^{51}-{\alpha }^{101}+3{\alpha }^{102}\right)\left(1{+}^4{C}_1\alpha {+}^5{C}_2{\alpha }^2+\dots +\mathrm{\infty }\right)$

$\begin{array}{l}{=}^{153}{C}_{150}-3{\times }^{102}{C}_{99}{-}^{52}{C}_{49}+3{\times }^{51}{C}_{48}\\ {=}^{153}{C}_3-3{\times }^{102}{C}_3{-}^{52}{C}_3+3{\times }^{51}{C}_3\\ =110556\end{array}$