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(C) The number of ways to get a total of $2m$ marks in $4$ papers,where each paper has a maximum of $m$ marks,is the coefficient of $x^{2m}$ in the ex

Vedclass · Accepted Answer

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

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(C) The number of ways to get a total of $2m$ marks in $4$ papers,where each paper has a maximum of $m$ marks,is the coefficient of $x^{2m}$ in the expansion of $(x^0 + x^1 + \dots + x^m)^4$.This is equal to the coefficient of $x^{2m}$ in $\left(\frac{1 - x^{m+1}}{1 - x}ight)^4$.$= 	ext{coefficient of } x^{2m} 	ext{ in } (1 - x^{m+1})^4 (1 - x)^{-4}$.$= 	ext{coefficient of } x^{2m} 	ext{ in } (1 - 4x^{m+1} + 6x^{2m+2} - \dots) \left(\sum_{r=0}^{\infty} \binom{r+3}{3} x^right)$.$= \binom{2m+3}{3} - 4 \binom{m+3-1}{3} = \binom{2m+3}{3} - 4 \binom{m+2}{3}$.$= \frac{(2m+3)(2m+2)(2m+1)}{6} - 4 \frac{(m+2)(m+1)m}{6}$.$= \frac{(m+1)}{6} [ (2m+3)(2)(2m+1) - 4m(m+2) ]$.$= \frac{(m+1)}{6} [ 2(4m^2 + 8m + 3) - 4m^2 - 8m ]$.$= \frac{(m+1)}{6} [ 8m^2 + 16m + 6 - 4m^2 - 8m ]$.$= \frac{(m+1)(4m^2 + 8m + 6)}{6} = \frac{(m+1)(2m^2 + 4m + 3)}{3}$.

$A$ person takes an examination consisting of four papers,each with a maximum of $m$ marks. The number of ways in which one can obtain a total of $2m$ marks is

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