Question

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

• A. ¹⁵³C₁₅₀
• B. 110556
• C. 3.¹⁰²C₉₉
• D. None of these

Answer 1

The correct option is B

110556

Aggregate of marks $=50\times 3+100=250$
$\therefore 60\%$ of the aggregate $=150$

Now the no. of ways of getting $150$ marks in aggregate $=$ coefficient of $x^{150}$ in ${(x^0+x^1+x^2+\dots ..+x^{50})}^3$ $(x^0+x^1+x^2+\dots \dots ..+x^{100})$

$=\text{coefficient of}{x}^{150}\text{in}{(1-x^{51})}^3(1-x^{101})(1-x{)}^{-4}$
$=\text{coefficient of}{x}^{150}\text{in}(1-3x^{51}+3x^{102}-x^{153})(1-x^{101})(1-x{)}^{-4}$
$=\text{coefficient of}{x}^{150}\text{in}(1-3x^{51}+3x^{102})(1-x^{101})(1-x{)}^{-4}$
$=\text{coefficient of}{x}^{150}\text{in}(1-3x^{51}-x^{101}+3x^{102})(1+{}^4{C}_{1}x+{}^5{C}_2{x}^2+\dots )$
$={}^{153}{C}_{150}-3\cdot {}^{102}{C}_{99}-{}^{52}{C}_{49}+3\cdot {}^{51}{C}_{48}$

$=110556.$