The correct option is B 110551
Let the marks scored by the candidate in four papers be x1,x2,x3,x4 then ,
x1+x2+x3+x4=150( i.e 60% of 250 is 150) where
0≤x1, x2, x3≤50 and 0≤x4≤100;
The number of solution of equation is same as the
coeff. of x150 in (1+x+x2⋯+x50)3(1+x+x2+⋯+x100)
=coeff. of x150 in (1−x511−x)3(1−x1011−x)
=coeff. of x150 in (1−x51)3(1−x101)(1−x)−4
=coeff. of x150 in (1−3x51+3x102−x153)(1−x101)×(1+ 4C3x+ 5C3x2+ 6C3x3⋯)
= 153C3−3× 102C3+3× 51C3− 52C3
=110551