Question

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

Answer 1

Answer: (C)

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