Question

A person opts 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. $\frac{1}{3}(m+1)(2m^2+4m+1)$
• C. $\frac{1}{3}(m+1)(2m^2+4m+3)$
• D. None of these

Answer 1

The correct option is C $\frac{1}{3}(m+1)(2m^2+4m+3)$

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 $(1-4x^{m+1}+6x^{2m+2}+...)(1+4x^{m+1}...\frac{(r+1)(r+2)(r+3)}{3!}{x}^r+...)$

$=\frac{(2m+1)(2m+2)(2m+3)}{6}-4m\frac{(m+1)(m+2)}{6}$

$=\frac{(m+1)(2m^2+4m+3)}{3}$