Question

In an examination, there are $4$ papers with a maximum of 'n' marks for each paper. Then the number of ways in which can one get $50\%$ marks is?

• A. $\frac{1}{3}(n+1)(2n^2+4n+3)$
• B. $\frac{1}{3}(n-1)(2n^2+4n+3)$
• C. $\frac{1}{3}(n+1)(2n^2-4n+3)$
• D. None of these

Answer 1

The correct option is A $\frac{1}{3}(n+1)(2n^2+4n+3)$

Let the $4$ papers be marked $x,y,z,w$

Let $m$ be the maximum marks

$0\le x,y,z,w\le m$

$x+y+z+w=2m$

Number of solutions : ${}^{2m+4-1}{C}_{4-1}{=}^{2m+3}{C}_3$

For, $x\ge m+1$

Let $t=x-(m+1)t\ge 0$

$x+y+z+w=2mt+y+z+w=m-1$

Number of solutions: ${}^{m-1+4-1}{C}_{4-1}{=}^{m+2}{C}_3$

For $y\ge m+1,z\ge m+1,w\ge m+1$

Final solution : ${}^{2m+3}{C}_3-4{\times }^{m+2}{C}_3$

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

Hence the correct answer is $\frac{1}{3}(m+1)(2m^2+4m+3)$