Question

In an examination, the maximum marks for each of three papers are $50$ each.

Maximum marks for the fourth paper are $100$.

The number of ways in which the candidate can score $60\%$ marks in the aggregate is

• A. $110556$
• B. $110500$
• C. $110356$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

Step-1: Solve for the required number of ways

The maximum marks that can be scored in all four papers$=50+50+50+100=250$

$60\%$ of the maximum aggregate is $\frac{60}{100}\times 250=150$ marks.

Let the marks scored in each paper be $x_1,x_2,x_3,x_4$ respectively

$0\le x_1,x_2,x_3\le 50$ and $0\le x_4\le 100$

To score $150$ marks in aggregate

$x_1+x_2+x_3+x_4=150$ subject to the above mentioned conditions

The number of ways of scoring $150$

$=$coefficient of $x^{150}$in${\left(1+x+x^2+...+x^{50}\right)}^3\left(1+x+x^2+...+x^{100}\right)$ [marks scored can be any out of $0$to $50$ for first three and $0$ to $100$ for the last paper]

$=$coefficient of $x^{150}$in ${\left(\frac{1-x^{51}}{1-x}\right)}^3\left(\frac{1-x^{101}}{1-x}\right)$ $\left[\because \left(1+x+x^2+...+x^n\right)=\frac{1-x^{n+1}}{1-x}\right]$

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

$=$coefficient of $x^{150}$ in $\left(1-3x^{51}+3x^{102}-x^{153}\right)\left(1-x^{101}\right){\left(1-x\right)}^{-4}\left[\because {\left(a-b\right)}^3=a^3-3a^{2}b+3a{b}^2-b^3\right]$

$=$ coefficient of $x^{150}$ in $(1-x^{153}-3x^{51}+3x^{102}-x^{101}-x^{254}+3x^{152}-3x^{203}){\left(1-x\right)}^{-4}$

Step-2: Solve further for the required number of ways

By using the binomial theorem we can write ${\left(1-x\right)}^{-n}=1+nx+\frac{n\left(n+1\right)}{2!}{x}^2+....$

Substituting the value of $n=-4$ we get ${(1-x)}^{-4}=1+{}^{4}C_{1}x+{}^5{C}_2{x}^2+....$

The number of ways of scoring $150$

$=$coefficient of $x^{150}$in $(1-x^{153}-3x^{51}+3x^{102}-x^{101}-x^{254}-3x^{152}-2x^{203})\left(1+{}^{4}C_{1}x+{}^5{C}_2{x}^2+....\right)$

Possible cases to get $x^{150}$ are $1\times x^{150}$ in second bracket, $-3x^{51}\times x^{99}$ in second bracket, $3x^{102}\times x^{48}$ in second bracket, $-x^{101}\times x^{49}$ in second bracket

The coefficient of $x^r$ in ${\left(1-x\right)}^{-n}$is ${}^{n+r-1}{C}_r$

Hence, the coefficient of $x^{150}$$=$${}^{4+150-1}{C}_{150}+\left(-3\right)·{}^{4+99-1}{C}_{99}+\left(3\right)·{}^{4+48-1}{C}_{48}+\left(-1\right)·{}^{4+49-1}{C}_{49}$

$=$${}^{153}{C}_{150}+\left(-3\right)·{}^{102}{C}_{99}+\left(3\right)·{}^{51}C_{48}+\left(-1\right)·{}^{52}{C}_{49}$

$=$$\frac{153!}{3!150!}-3\times \frac{102!}{99!3!}+\left(3\right)\frac{51!}{48!3!}-\frac{52!}{49!3!}$ $\left[\because {}^{n}C_r=\frac{n!}{r!\left(n-r\right)!}\right]$

$=$$585276-3\left(171700\right)+3\left(20825\right)-22100$

$\Rightarrow$ The coefficient of $x^{150}$$=$$110551$

Thus the number of ways to score $60\%$ marks aggregate are $110551$.

Hence option(D) i.e. None of these is the correct answer.