Question

A person goes in for an examination in which there are four papers with a maximum of 20 marks from each paper. The number of ways in which one can get 40 marks is_______.

• A.
• B. 6167
• C. 6181
• D. 7000

Answer 1

The correct option is C

6181

$p_1+p_2+p_3+p_4=40$

Where $0\le p_1,p_2,p_3,p_4\le 20$

The required number

= coefficient of $x^{40}$ in $(x^0+x^1+..........+x^{20}{)}^4$

= coefficient of $x^{40}$ in ${\left(\frac{1-x^{21}}{1-x}\right)}^4$

= coefficient of $x^{40}$ in $(1-x^{21}{)}^4(1-x{)}^{-4}$

= coefficient of $x^{40}$ in $(1-4x^{21}+6x^{42}+............)(1+4x+...........+\frac{(r+1)(r+2)(r+3)}{3!}{x}^r+.................)$

=$\frac{(40+1)(40+2)(40+3)}{6}-80\frac{(20+1)(20+2)}{6}$

= 12341 - 6160

= 6181