Question

A man goes in for an examination in which there are $4$ papers with a maximum of $10$ marks for each paper. The number of ways of getting $20$ marks on the whole is

• A. $750$
• B. $860$
• C. $881$
• D. $891$

Answer 1

Answer: (D)

$891$

If we take the marks in each test to be $m$.
The required answer is the coefficient of $x^{2m}$ in ${(x^0+x^1+...x^m)}^4={\left(\frac{1-x^{m+1}}{1-x}\right)}^4={(1-x^{m+1})}^4.{(1-x)}^{-4}$
$={(1-x^{m+1})}^4.(1{+}_1^{4}C.x{+}_2^{5}C.x^2{+}_3^{6}C.x^3+..{.}_{(m-1)}^{(m+2)}C.x^{m-1}+....{+}_{2m}^{(2m+3)}C.x^{2m}+.....\infty )$
${=}_{2m}^{(2m+3)}C-4{\times }_{(m-1)}^{(m+2)}C$
${=}_3^{(2m+3)}C-4{\times }_3^{(m+2)}C$
$=\frac{(2m+3)(2m+2)(2m+1)}{6}-4\times \frac{(m+2)(m+1)\left(m\right)}{6}=\frac{2(2m+3)(m+1)(2m+1)}{6}-4\times \frac{(m+2)(m+1)\left(m\right)}{6}$
$=\frac{(m+1)}{3}\times \frac{(2m+3)(2m+1)-2m(m+2)}{3}=\frac{(m+1)(2m^2+4m+3)}{3}$
Here, $m=10$
Thus, after substituting, the answer is $\frac{11\times 243}{3}=891$