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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+4-1C4-1=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
t+y+z+w=m -1
solution of this equation = m-1+4-1C4-1=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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