Question

In an examination, the maximum marks for first three papers is $n$, and for the fourth paper is $2n$. The number of ways in which a student can secure $3n$ marks is

• A. $\frac{1}{6}(n-1)(5n^2+10n+6)$
• B. $\frac{1}{6}(n+1)(5n^2+10n+6)$
• C. $\frac{1}{6}(n+1)(5n^2+n+6)$
• D. $Noneofthese$

Answer 1

The correct option is D $Noneofthese$

This is equivalent to finding the coefficient of $x^{3n}$ in ${(1+x+x^2+...+x^n)}^3.(1+x+x^2+...+x^{2n})$$\Rightarrow \frac{{(1-x^{n+1})}^3(1-x^{2n+1})}{{(1-x)}^3(1-x)}={(1-x^{n+1})}^3(1-x^{2n+1}){(1-x)}^{-4}$
$\Rightarrow (1-3x^{n+1}+3x^{2n+2}-x^{3n+3})(1-x^{2n+1})(1{+}_1^{4}Cx{+}_2^{5}C{x}^2+...{+}_n^{(n+3)}C{x}^n+...\infty )\Rightarrow (1-x^{2n+1}+3x^{2n+2}-3x^{n+1}).(1{+}_1^{4}Cx+...{+}_{(n-2)}^{(n+1)}C{x}^{n-2}{+}_{(n-1)}^{(n+2)}C{x}^{n-1}{+}_{(2n-1)}^{(2n+2)}C{x}^{2n-1}{+}_{\left(3n\right)}^{(3n+3)}C{x}^{3n}+...\infty )\Rightarrow (1-x^{2n+1}+3x^{2n+2}-3x^{n+1}).(1{+}_1^{4}Cx+...{+}_{\left(3\right)}^{(n+1)}C{x}^{n-2}{+}_{\left(3\right)}^{(n+2)}C{x}^{n-1}{+}_{\left(3\right)}^{(2n+2)}C{x}^{2n-1}{+}_{\left(3\right)}^{(3n+3)}C{x}^{3n}+...\infty ){\Rightarrow }_{\left(3\right)}^{(3n+3)}C-{3.}_{\left(3\right)}^{(2n+2)}C+{3.}_{\left(3\right)}^{(n+1)}C-{.}_{\left(3\right)}^{(n+2)}C\Rightarrow \frac{(3n+3)(3n+2)(3n+1)}{6}-\frac{3.(2n+2)(2n+1)\left(2n\right)}{6}+\frac{3.(n+1)n(n-1)}{6}-\frac{(n+2)(n+1)n}{6}\Rightarrow \frac{3.(n+1)(3n+2)(3n+1)}{6}-\frac{n(n+1)}{6}.\left(12\right(2n+1)-3(n-1)+(n+2\left)\right)\Rightarrow \frac{3.(n+1)(3n+2)(3n+1)}{6}-\frac{n(n+1)}{6}.(22n+17)\Rightarrow \frac{(n+1)}{6}.(5n^2+10n+6)$
Hence, (d) is correct.