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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

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

Vedclass · Answer

(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 equivalent to the coefficient of $x^{2m}$ in $\left( \frac{1 - x^{m+1}}{1 - x} ight)^4 = (1 - x^{m+1})^4 (1 - x)^{-4}$.Expanding this,we get $(1 - 4x^{m+1} + 6x^{2m+2} - \dots) 	imes \sum_{r=0}^{\infty} \binom{r+3}{3} x^r$.The coefficient of $x^{2m}$ is obtained by taking the term $1 	imes \binom{2m+3}{3}$ and the term $-4x^{m+1} 	imes \binom{m+2}{3}$.Number of ways = $\binom{2m+3}{3} - 4 \binom{m+2}{3} = \frac{(2m+3)(2m+2)(2m+1)}{6} - 4 \frac{(m+2)(m+1)m}{6}$.Simplifying this expression: $\frac{(m+1)}{6} [ (2m+3)(2)(2m+1) - 4m(m+2) ] = \frac{(m+1)}{6} [ 2(8m^2 + 8m + 3) - 4m^2 - 8m ] = \frac{(m+1)}{6} [ 16m^2 + 16m + 6 - 4m^2 - 8m ] = \frac{(m+1)(12m^2 + 8m + 6)}{6} = \frac{(m+1)(6m^2 + 4m + 3)}{3}$.Wait,re-evaluating the expansion: $\binom{2m+3}{3} - 4\binom{m+2}{3} = \frac{(2m+3)(2m+2)(2m+1) - 4(m+2)(m+1)m}{6} = \frac{(m+1)[(2m+3)(2)(2m+1) - 4m(m+2)]}{6} = \frac{(m+1)[2(4m^2+8m+3) - 4m^2-8m]}{6} = \frac{(m+1)[8m^2+16m+6-4m^2-8m]}{6} = \frac{(m+1)(4m^2+8m+6)}{6} = \frac{(m+1)(2m^2+4m+3)}{3}$.

$A$ 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

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