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

The correct option is C

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

Explanation for the correct option:

Calculate the required number of ways :

As per the given condition, the required number of ways

$=$coefficient of $x^{2m}$in ${(x_o+x_1+............+x_m)}^4$

$=$coefficient of $x^{2m}$in ${\left[\frac{(1-x^{m+1})}{(1-x)}\right]}^4$

$=$coefficient of $x^{2m}$in $\left[{(1-x^{m+1})}^4\times (1-x{)}^{-4}\right]$

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

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

Hence, the correct option is $\left(C\right)$.