In an examination, the maximum mark for each of the three papers is 50 and the maximum mark for the fourth paper is 100. Find the number of ways in which the candidate can score 605 marks in aggregate.

Explanation

Aggregate of marks`=50xx3+100=250`
`therefore60%` of the aggregate=`(60)/(100)xx250=150`
Let the mark scored by the candidate in four papers be `x_(1),x_(2), x_(3) and x_4`. Here, the number of required ways will be equal to the number of sols of `x_(1)+x_(2)+x_(3)+x_(4)=150` i.e.,
`0 le x_(1),x_(2),x_(3) le 50 and xlex_(4) le 100`.
Since the upper limit is `100 lt `required sum (150). ,brgt The number of solutions of the equation is equal to ccoefficient of `alpha^(150)` in the expansion of
`(alpha^(0)+alpha^(1)+alpha^(2)+ . ..+alpha^(50))^(3)(alpha^(0)+alpha^(1)+alpha^(2)+ . . .+alpha^(100))`
=Coefficient of `alpha^(150)` in the expansion of
`(1-alpha^(51))^(3)(1-alpha^(10))(1-alpha)^(-4)`
=Coefficient of `alpha^(150)` in the expansion of
`(1-3alpha^(51)+3alpha^(102))(1-alpha^(101))(1+.^(4)C_(1)alpha+.^(5)C_(2)alpha^(2)+ . . .+oo)`
=Coefficient of `alpha^(150)` in the expansion of
`(1-3alpha^(51)+3alpha^(102))(1-alpha^(101))(1+.^(4)C_(1)alpha+.^(5)C_(2)alpha^(2)+ . . .+oo)`
=Coefficient of `alpha^(150)` in expansion of
`(1-3alpha^(51)+3alpha^(101)+3alpha^(102))(1+.^(4)C_(1)alpha+.^(5)C_(2)alpha^(2)+ . ..+oo)`
`=.^(153)C_(150)-3xx.^(102)C_(99)-.^(52)C_(49)+3xx.^(51)C_(48)`
`=.^(153)C_(3)-3xx.^(102)C_(3)-.^(52)C_(3)+3xx.^(51)C_(3)`
`=110556`