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