In an examination, the maximum marks for each of three subjects is $n$ and that for the fourth subject is $2…
In an examination, the maximum marks for each of three subjects is $n$ and that for the fourth subject is $2 n$. The number of ways in which candidates can get $3 n$ marks is
$\frac{1}{6}(n+1)^2\left(5 n^2+10 n+6\right)^2$
$\frac{1}{6}(n+1)\left(5 n^2+10 n+6\right)^2$
$\frac{1}{6}(n+1)^2\left(5 n^2+10 n+6\right)$
$\frac{1}{6}(n+1)\left(5 n^2+10 n+6\right)$
Solution
Total marks $=$ Marks for first 3 papers + Marks
for fourth paper $=3 n+2 n=5 n$
Candidate needs to get $3 n$ marks.
Let $x_1, x_2, x_3, x_4$ be the marks of candidate in I, II, III and IV paper, respectively.
Then, $x_1+x_2+x_3+x_4$
$=3 n\left(0 \leq x_1, x_2, x_3, \leq n\right.$ and $\left.0 \leq x_4 \leq 2 n\right)$
Using multinomial theorem,
Number of ways $=$ Coefficient of $x^{3 n}$ in
$\underbrace{\left(1+x+\ldots .+x^n\right)^3}_{\text {GP series }} \underbrace{\left(1+x+\ldots x^{2 n}\right)}_{\text {GP Series }}$
$\begin{aligned} & \Rightarrow 1-x^{3(n+1)}-3 x^{n+1}\left(1-x^{n+1}\right)\left(1-x^{2 n+1}\right)(1-x)^{-4} \\ & \Rightarrow\left(1-x^{3(n+1)}-3 x^{n+1}+3 x^{2 n+2}\right)\left(1-x^{2 n+1}\right)(1-x)^{-4}\end{aligned}$
$\begin{aligned} \Rightarrow\left(1-x^{3(n+1)}-3 x^{n+1}+\right. & 3 x^{2 n+2}-x^{2 n+1}+x^{5 n+4} \\ & \left.+3 x^{3 n+2}-3 x^{4 n+3}(1-x)^{-4}\right)\end{aligned}$
Finding the coefficient of $x^{3 n}$ using binomial expansion, we get
${ }^{3 n+3} C_3-3^{2 n+2} C_3+3^{n+2} C_3-{ }^{n+3} C_3$
$=\frac{1}{6}(n+1)\left(5 n^2+10 n+6\right)$