The correct option is B 21C7
Let xi denote the marks assigned to the ith question. Then
x1+x2+x3+⋯+x7+x8=30
where i = 1, 2, 3, . . ., 8 and xi≥2
and y1+y2+⋯+y7+y8=14
where yi=xi−2, i = 1, 2, 3 . . . .
∴ The total number of solutions of this equation is
14+8−1C8−1=21C7