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(D) Let $x_1, x_2, x_3$ be the marks in the first three subjects $(0 \leq x_i \leq n)$ and $x_4$ be the marks in the fourth subject $(0 \leq x_4 \leq

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$\frac{1}{6}(n+1)(5n^2+10n+6)$

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(D) Let $x_1, x_2, x_3$ be the marks in the first three subjects $(0 \leq x_i \leq n)$ and $x_4$ be the marks in the fourth subject $(0 \leq x_4 \leq 2n)$.We need to find the number of integer solutions to $x_1 + x_2 + x_3 + x_4 = 3n$.This is the coefficient of $x^{3n}$ in the expansion of $(1+x+\dots+x^n)^3(1+x+\dots+x^{2n})$.This expression is equal to $\left(\frac{1-x^{n+1}}{1-x}\right)^3 \left(\frac{1-x^{2n+1}}{1-x}\right) = (1-x^{n+1})^3(1-x^{2n+1})(1-x)^{-4}$.Expanding this,we get $(1 - 3x^{n+1} + 3x^{2n+2} - x^{3n+3})(1 - x^{2n+1})(1-x)^{-4}$.$= (1 - 3x^{n+1} + 3x^{2n+2} - x^{3n+3} - x^{2n+1} + 3x^{3n+2} - 3x^{4n+3} + x^{5n+4})(1-x)^{-4}$.We need the coefficient of $x^{3n}$ in this product.Using the expansion $(1-x)^{-4} = \sum_{r=0}^{\infty} \binom{r+3}{3} x^r$,the coefficient is:$\binom{3n+3}{3} - 3\binom{2n+2}{3} - \binom{n+2}{3} + 3\binom{n-1}{3}$ (where $\binom{n}{k} = 0$ if $n Calculating this,we get $\frac{(3n+3)(3n+2)(3n+1)}{6} - 3\frac{(2n+2)(2n+1)(2n)}{6} - \frac{(n+2)(n+1)n}{6} + 3\frac{(n-1)(n-2)(n-3)}{6}$.Simplifying this expression yields $\frac{1}{6}(n+1)(5n^2+10n+6)$.

In an examination,the maximum marks for each of three subjects is $n$ and that for the fourth subject is $2n$. The number of ways in which a candidate can get $3n$ marks is

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