Question

# There are three papers of $100$ marks each in an examination. In how many ways a student gets $150$ marks such that he gets at least $60$ percent in two papers.A) $1480$ B) $1488$ C) $1520$ D) $1526$

Hint: In this problem, use a combination method to solve it. In the question student gets $150$ marks such that he gets at least $60$ percent in two papers. So we split $150$ according to it. And perform questions using a combination method also in a number of ways according to the given question.
Formula: $^n{C_r} = \dfrac{{n!}}{{r! \times \left( {n - r} \right)!}}$

Complete step by step solution: Given that-
Student gets $150$ marks such that he gets at least $60$ percent in two papers.
$150 = \left( {60 + 60 + 30} \right)$
Because $60$ percent in two papers and remaining are$30$ to complete$150$.
Number of ways$=$ coefficient of ${x^{150}}$ is
$\left\{ {{{\left( {{x^{60}} + {x^{61}} + .......{x^{100}}} \right)}^2}\left( {1 + x + {x^{2 + }}.........{x^{30}}} \right)} \right\}$
Coefficient of ${x^{30}}$ is
$\left\{ {{{\left( {1 + x + {x^{2 + }}.........{x^{40}}} \right)}^2}\left( {1 + x + {x^2} + .........{x^{30}}} \right)} \right\}$
Coefficient of ${x^{30}}$ is
${\left( {\dfrac{{1 - {x^{41}}}}{{1 - x}}} \right)^2}\left( {\dfrac{{1 - {x^{31}}}}{{1 - x}}} \right) \\ \Rightarrow {\left( {1 - x} \right)^{ - 3}} \\ \Rightarrow {x^{30 + 3 - 1}}{C_{3 - 1}} \\ { \Rightarrow ^{32}}{C_2} \\$
Thus the student gets $60$ percent marks in first two papers to get $150$ marks
Number of ways ${ = ^{32}}{C_2}$
But the two paper at least $60$ percent, can be chosen out of $3$ papers
Number of ways${ = ^3}{C_2}$
Required number of ways ${ = ^3}{C_2}{ \times ^{32}}{C_2}$
To solve above combination use formula
$^n{C_r} = \dfrac{{n!}}{{r! \times \left( {n - r} \right)!}}$
${ = ^3}{C_2}{ \times ^{32}}{C_2}$
$\dfrac{{3!}}{{2! \times \left( {2 - 1} \right)!}} \times \dfrac{{32!}}{{2!\left( {32 - 2} \right)!}} \\ \Rightarrow \dfrac{{3!}}{{2! \times 1!}} \times \dfrac{{32!}}{{2! \times 30!}} \\ \Rightarrow \dfrac{{3 \times 2 \times 1}}{{2 \times 1 \times 1}} \times \dfrac{{32 \times 31 \times 30 \times 29........2 \times 1}}{{2 \times 1 \times 30 \times 29.....2 \times 1}} \\ \Rightarrow 3 \times 496 \\ \Rightarrow 1488 \\$ To solve it we get:
Number of ways $= 1488$

Hence option B is the correct answer.

Note: First we have the total number of ways according to the question. By making these combinations we have a required number of total ways. After that we solve the number of ways by using the formula. This we get the total number of ways a student gets $150$ marks such that he gets at least $60$ percent in two papers. Hence the answer is 1488 .