
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) $1526$
Answer
482.4k+ views
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 .
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 .
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Express the following as a fraction and simplify a class 7 maths CBSE

The length and width of a rectangle are in ratio of class 7 maths CBSE

The ratio of the income to the expenditure of a family class 7 maths CBSE

How do you write 025 million in scientific notatio class 7 maths CBSE

How do you convert 295 meters per second to kilometers class 7 maths CBSE

Trending doubts
List some examples of Rabi and Kharif crops class 8 biology CBSE

State the differences between manure and fertilize class 8 biology CBSE

Public administration is concerned with the administration class 8 social science CBSE

What led to the incident of Bloody Sunday in Russia class 8 social science CBSE

What is the tagline of Swachh Bharat Abhiyaan A Sabka class 8 social studies CBSE

State whether true or false Every rhombus is a square class 8 maths CBSE
