• # question_answer 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 $2m$ marks is A) $^{2m+3}{{C}_{3}}$ B) $\frac{1}{3}(m+1)(2{{m}^{2}}+4m+1)$ C) $\frac{1}{3}(m+1)(2{{m}^{2}}+4m+3)$ D) None of these

The required number = coefficient of ${{x}^{2m}}$ in ${{({{x}^{0}}+{{x}^{1}}+......+{{x}^{m}})}^{4}}$ = coefficient of ${{x}^{2m}}$ in ${{\left( \frac{1-{{x}^{m+1}}}{1-x} \right)}^{4}}$ = coefficient of ${{x}^{2m}}$ in ${{(1-{{x}^{m+1}})}^{4}}{{(1-x)}^{-4}}$ = coefficient of ${{x}^{2m}}$ in $(1-4{{x}^{m+1}}+6{{x}^{2m+2}}+......)$                                     $\left( 1+4x+......+\frac{(r+1)(r+2)(r+3)}{3\ !}{{x}^{r}}+.... \right)$ $=\frac{(2m+1)(2m+2)(2m+3)}{6}-4m\frac{(m+1)(m+2)}{6}$ $=\frac{(m+1)(2{{m}^{2}}+4m+3)}{3}$.