The correct option is C 13
Let E=(x−1)(x2−2)(x3−3)⋯(x20−20)
⇒E=x(1−1x)⋅x2(1−2x2)⋅x3(1−3x3)⋯x20(1−20x20)
⇒E=x210(1−1x)(1−2x2)⋯(1−20x20)
Now, coefficient of x203 in E is coefficient of x−7 in (1−1x)(1−2x2)⋯(1−20x20)
Coefficient of x–7
=−7+1⋅6+2⋅5+3⋅4−1⋅2⋅4
=−7+6+10+12−8
=13