Question

In an examination the maximum marks for each of the three papers are $50$ each. maximum marks for the fourth paper $100.$ The number of ways in which the candidate can score $60$% marks in aggregate is

• A. $110256$
• B. $110456$
• C. $110556$
• D. None of these

Answer 1

The correct option is C $110556$

The candidate must score 150 marks.

Thus required number

= coefficient of $x^{150}$ in ${(1+x......+x^{50})}^3(1+x......+x^{100})$

= coefficient of $x^{150}$ in ${\left(\frac{1-x^{51}}{1-x}\right)}^3\left(\frac{1-x^{101}}{1-x}\right)$

= coefficient of $x^{150}$ in ${(1-x^{51})}^3{(1-x^{101})(1-x)}^{-4}$

= coefficient of $x^{150}$ in $(1-3x^{51}+3x^{102}-x^{153})(1-x^{101}){(1-x)}^{-4}$ (leaving terms having power of x greater than 150).

= coefficient of $x^{150}$ in ${(1-x)}^{-4}$ - 3 times coefficient of $x^{99}$ in ${(1-x)}^{-4}$ + 3 times coefficient of $x^{48}$ in ${(1-x)}^{-4}$ - coefficient of $x^{49}$ in ${(1-x)}^{-4}$

$={{}^{153}C}_{150}-3{{}^{102}C}_{99}+3{{}^{51}C}_{48}-{{}^{52}C}_{49}=110556$