q37

Saumyojit

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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)
 
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
Now , we have to find coefficient of x^150
Using the generating function [imath]{\left( {\sum\limits_{k = 0}^{50} {{x^k}} } \right)^3}\left( {\sum\limits_{k = 0}^{100} {{x^k}} } \right)[/imath], its expansion contains the term [imath]110551x^{150}[/imath] SEE HERE
 
Using the generating function (∑k=050xk)3(∑k=0100xk){\left( {\sum\limits_{k = 0}^{50} {{x^k}} } \right)^3}\left( {\sum\limits_{k = 0}^{100} {{x^k}} } \right)(k=0∑50xk)3(k=0∑100xk), its expansion contains the term 110551x150110551x^{150}110551x150
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=coefficient of x^150 in (1+x+...+x50)^3 (1+x+...+x100)
= coefficient of x^150 in (1−x^51)^3*(1−x^101)*

(1 −x)^-4

=coefficient of x^150 in

(1−3x^51+3x^102−x^153)(1−x^101)(1−x)^−4
[leaving terms containing powers of x greater than 150]



Then?
 
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