Question

In an examination, the maximum marks for each of three papers are $50$ each. Maximum marks for the $4^{th}$ paper is $100$. Find the number of ways in which the candidate can score $60$% marks in the aggregate.

• A. $110550$
• B. $110551$
• C. $122050$
• D. $122051$

Answer 1

Answer: (B)

$110551$

The candidate must score $150$ marks.
$\therefore$ Required number
$=$coefficient of $x^{150}$ in ${(1+x+...+x^{50})}^3(1+x+...+x^{100})$
$=$ coefficient of $x^{150}$ in ${\left(\frac{1-x^{51}}{1-x}\right)}^3\frac{1-x^{101}}{1-x}$
$=$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 containing powers of x greater than $150$]
$=$coefficient of $x^{150}$ in ${(1-x)}^{-4}-3$.
$=$coefficient of $x^{99}$ in ${(1-x)}^{-4}+3$coefficient of $x^{48}$ in ${(1-x)}^{-4}$coefficient of $x^{49}$in${(1-x)}^{-4}$
${=}^{153}{C}_{150}-{3.}^{102}{C}_{99}+{3.}^{51}{C}_{48}{-}^{52}{C}_{49}$
$=\frac{\mathrm{153.152.151}}{6}-3.\frac{\mathrm{102.101.100}}{6}+3.\frac{\mathrm{51.50.49}}{6}-\frac{\mathrm{52.51.50}}{6}$
$=110551$