In an examination, the maximum marks for each of three papers are 50 each. Maximum marks for the 4th paper is 100. Find the number of ways in which the candidate can score 60% marks in the aggregate.
CAndidate needs to score 150 .
x1+x2+x3+x4 =150
(0+ x^0+x^1+...+x^50) + ( 0+ x^0+x^1+...+x^50)_+(0+ x^0+x^1+...+x^50) + (0+ x^0+x^1+...+x^100 )
(x^0+x^1+...+x^50)^ 3 + (x^0+x^1+...+x^100 )
As it is in GP, sum = [ 1(1- x ^51) ] ^3 / (1-x) ^3
sum = [1(1-x ^101) ] /(1-x)
:: [ 1(1- x ^51) ]^3 /(1-x) ^3 + [1(1-x ^101) ] /(1-x)
(1-x^51)^3 * (1-x)^-3 + (1-x^101) * (1-x)^-1
Now , we have to find coefficient of x^150
How to do? we have to use multinomial THEOREM which is (n+r-1) C r which gives coefficient of x^150 in the expansion of (1-x)^ (-n)