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Question

Answers

A) $1480$

B) $1488$

C) $1520$

D) $1526$

Answer
Verified

Formula: $^n{C_r} = \dfrac{{n!}}{{r! \times \left( {n - r} \right)!}}$

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$