a man goes for an examination in which there are 4 papers with maximum marks'm' for each paper then no of ways of getting 2m marks:ans. (m+1)(2m^2+4m+3)/3

Dear shefali

let the markd in the 4 papers are x,y,z,w

 where 0≤x,y,z,w≤m

  x+y+z+w=2m

now our task is to find out the intiger solution of this equation in given condition.

number of solution is   = ^2m+4-1C_4-1=^2m+3C₃

but this solution also include those solution in which any variable is greater than m

so we have to subtact those solution

for  x ≥ m+1 .

let  t=x-(m+1)     t≥o

x+y+z+w=2m

t+y+z+w=m -1

solution of this equation = ^m-1+4-1C_4-1=^m+2C₃

similerly for y ≥ m+1,z ≥ m+1,w ≥ m+1

 so final solution =^2m+3C₃ -4*^m+2C₃

                        =(m+1)(2m^2+4m+3)/3

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