## Question

There are three papers of *100* marks each in an examination. Then the no. of ways can a student get *150* marks such that he gets atleast *60%* in two papers

### Solution

^{3}C_{2} × ^{32}C_{2}

Therefore,

Sum of marks obtained = 150

The required number of ways

= No, of integral solutions of (i)

= Coefficient of *x*^{150 }in {*x*^{60 }+ *x*^{61 }+ …+*x*^{100})^{2} (1+ *x* + *x*^{2}+ …+ *x*^{30})}

= Coefficient of *x*^{30} in {(1 + *x* + … + *x*^{40})^{2 }(1 + *x* + … + *x*^{30})}

= Coefficient of *x*^{30} in (1 – *x*)^{-3}

= ^{30 + 3 – }^{1 }C_{3 – 1 }= ^{32}C_{2}.

Thus, the student gets atleast *60%* marks in first two papers to get 150 marks as total in ^{32}C_{2} ways. But the two papers, of atleast

*60%* marks, can be chosen out of three papers in ^{3}C_{2} ways.

Hence the required number of ways = ^{3}C_{2} × ^{32}C_{2}.

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