Question

There are three papers of $100$ marks each in an examination. In how many ways can a student get $150$ marks such that he gets at least $60$ in two papers?

• A. $1480$
• B. $1488$
• C. $1520$
• D. $1526$

Answer 1

The correct option is B $1488$

Suppose the student gets at least $60$ marks in first two papers, then he just gets at most $30$ marks in the third paper to make a total of $150$ marks

So, the required numbers of ways = coefficient of $x^{150}$ in ${(x^{60}+x^{61}+...+x^{100})}^2(1+x+x^2+...+x^{30})$

$=$ coefficient of $x^{30}$ in $\left({(1+x+...+x^{40})}^2(1+x+...x^{30})\right)$

$=$ coefficient of $x^{30}$ in ${\left(\frac{1-x^{41}}{1-x}\right)}^2\left(\frac{1-x^{31}}{1-x}\right)$

$=$ coefficient of $x^{30}$ in ${(1-x)}^{-3}$

${=}^{30+3-1}{C}_{3-1}{=}^{32}{C}_2$

Thus the students get at least $60$ marks in the first two papers to get 150 marks as total in ${}^{32}{C}_2$ ways. But the two papers at least $60$, can be chosen out of 3 papers in ${}^3{C}_2$ ways.

Hence,the required number of ways ${}^3{C}_2{\times }^{32}{C}_2=1488$