A person goes in for an examination in which there are four papers with a maximum of m marks from each paper. The number of ways in which one can get 2 m marks, is 

The required number 

= Coeff. of $x^{2m}$ in ${\left(x^0+x^1+\dots +x^m\right)}^3$

$=\text{ Coeff. of }{x}^{2m}\text{ in }{\left(\frac{1-x^{m+1}}{1-x}\right)}^3$

$=\text{ Coeff of }{x}^{2m}\text{ in }{\left(1-x^{m+1}\right)}^3(1-x{)}^{-4}$

$=\text{ Coeff. of }{x}^{2m}\text{ in }\left\{\left(1-4x^{m+1}+6x^{2m-2}+\dots \right)\right.$

$\times \left(1+4x+\dots +\frac{(r+1)(r+2)(r+3)}{3!}{x}^r+\dots \right)$

$\begin{array}{l}=\frac{(2m+1)(2m+2)(2m+3)}{6}-4m\frac{(m+1)(m+2)}{6}\\ =\frac{(m+1)\left(2m^2+4m+3\right)}{3}\end{array}$