Question

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

• A. $125640$
• B. $110551$
• C. $97860$
• D. None of these

Answer 1

The correct option is B $110551$

Let the marks scored by the candidate in four papers be $x_1,x_2,x_3,x_4\text{then},$
$x_1+x_2+x_3+x_4=150$( i.e $60\%$ of $250$ is $150)$ where
$0\le x_1,x_2,x_3\le 50\text{and}0\le x_4\le 100;$
The number of solution of equation is same as the
$\text{coeff. of}{x}^{150}\text{in}(1+x+x^2\cdots +x^{50}{)}^3(1+x+x^2+\cdots +x^{100})$
$=\text{coeff. of}{x}^{150}\text{in}{\left(\frac{1-x^{51}}{1-x}\right)}^3\left(\frac{1-x^{101}}{1-x}\right)$
$=\text{coeff. of}{x}^{150}\text{in}(1-x^{51}{)}^3(1-x^{101}\left)\right(1-x{)}^{-4}$
$=\text{coeff. of}{x}^{150}\text{in}(1-3x^{51}+3x^{102}-x^{153})(1-x^{101})\times (1+{}^4{C}_{3}x+{}^5{C}_3{x}^2+{}^6{C}_3{x}^3\cdots )$
$={}^{153}{C}_3-3\times {}^{102}{C}_3+3\times {}^{51}{C}_3-{}^{52}{C}_3$
$=110551$